Understanding limit of exponents

[DarthMatter]

But the limit doesn't need to exist because the limit is already finished now. The limit was already evaluated on g(x) and so it doesn't exist anymore. That means that there is no limit operating on h(x) since it was operated on g(x) and since then it vanished.

If you are speaking about the $lim_{x\rightarrow a} h(g(x))$, then where does it say that $lim_{x\rightarrow a} h(g(x))=h(lim_{x\rightarrow a}g(x))$? It's not a definition, and the theorem does not hold because h is not continuous. Maybe you should first try to understand what continuous means in the $\varepsilon$-$\delta$-formalism (it's really good, sophisticated and worth the effort) and come back to composite functions later. :) The problem is that without a feel why this $\varepsilon$ and this $\delta$ is chosen you cannot really grasp the concepts 'behind the scenes'.

 

[andyrk]

If you are speaking about the lim_{x\rightarrow a} h(g(x)), then where does it say that lim_{x\rightarrow a} h(g(x))=h(lim_{x\rightarrow a}g(x))? It's not a definition, and the theorem does not hold because h is not continuous. Maybe you should first try to understand what continuous means in the ε-δ-formalism (it's really good, sophisticated and worth the effort) and come back to composite functions later. :) The problem is that without a feel why this ε and this δ is chosen you cannot really grasp the concepts 'behind the scenes'.

I could never make any sense of the ε-δ proof. And I don't think I will be able to until some miracle happens. :(

 

[Stephen Tashi]

That's because my course material says to directly substitute the value to which x is approaching, in the function (g(x) in this case).

If your course material says you may do that without knowing h() is continuous at L then your course material is wrong. The "composite function limit theorem" tells you when you may make that substitution.

A limit is a number to which y = f(x) approaches as x approaches to some arbitrary number say a. So, it is a number which the function approaches but never reaches [/itex]

Don't say "never reaches". You mean "may or may not reach".

but I never said that it has to reach there. I just said that wherever it is approaching but never reaching is what the limit evaluates to.

Ok, but we aren't going to get to the actual definition of limit by talking about "approaching" or "reaching" or "not reaching". The epsilon-delta definition approach to this question has been explained in detail by Fredrik. If you aren't at the point where you can understand it, then the best we can do is examples. You intuition seems to insist that lim_{x \rightarrow a} h(g(x)) must be the same thing as h( lim_{x\rightarrow a} g(x)). We covered examples where this is not the case.

 

[andyrk]

If your course material says you may do that without knowing h() is continuous at L then your course material is wrong. The "composite function limit theorem" tells you when you may make that substitution.

My course material says nothing about composite limit theorem. It says to substitute only in evaluating simple limits not composite. And infact I never encountered anywhere in all my problem sets or course material any reference to composite limits. I just encountered- lim_{x\rightarrow a} e^{f(x)} being equated to e^{lim_{x\rightarrow a} f(x)}. Does this require the understanding of composite function limit theorem? My course material is just advanced high school calculus (maybe that's the reason it doesn't cover it).

If you aren't at the point where you can understand it, then the best we can do is examples. You intuition seems to insist that lim_{x \rightarrow a} h(g(x)) must be the same thing as h( lim_{x\rightarrow a} g(x)) . We covered examples where this is not the case.

Then I think I would just go with my intuition because that way things are less complicated and also, I think I am going much beyond what my course material asks for.

 

[DarthMatter]

I could never make any sense of the ε-δ proof. And I don't think I will be able to until some miracle happens. :(

You can think of it as a bet you make with a friend: Your friend has a function f (f because it is his favorite function) and some point c. Now the bet goes like this: Your friend bets to any number $\epsilon>0$ you tell him he can give you some $\delta>0$ such that for all points x' which are not farer from c than $\delta$ (which means $|x'-c| < \delta$) the function values f(x') and f(c) don't differ by more than $\varepsilon$ (this means $|f(x')-f(c)| < \varepsilon$).

So, if you find any point $x'$ for which $|x'-c|<\delta$ AND $|f(x')-f(c)|>\varepsilon$ your friend has failed to deliver a fitting $\delta > 0$ and you win the bet. You can try as often as you like with any $\varepsilon>0$ you want, your friend always has to give you a fitting $\delta>0$, and it is enough for you to find a single $x'$ with $|f(x')-f(c)|>\varepsilon$ for you to win. So you are in quite a good position.

However, if you can never find such an $\varepsilon$ and $x'$ which breaks the deal, your friend wins, shouts out the battlecry 'f IS CONTINUOUS AT c!' and pours an icebucket over your head. If on the other hand you find an $\varepsilon$ with a fitting $x'$, you may shout 'f IS NOT CONTINUOUS AT c!' victoriously and push him into a bathtub of vanilla pudding. So it's always quite a mess, but also a good day for mathematics.

 

[Stephen Tashi]

My course material says nothing about composite limit theorem. It says to substitute only in evaluating simple limits not composite. And infact I never encountered anywhere in all my problem sets or course material any reference to composite limits. I just encountered- lim_{x\rightarrow a} e^{f(x)} being equated to e^{lim_{x\rightarrow a} f(x)}. Does this require the understanding of composite function limit theorem? My course material is just advanced high school calculus (maybe that's the reason it doesn't cover it).

It's possible your material doesn't cover it, but most modern U.S. textbooks would at least have a statement in the "fine print" about when you can apply that procedure and when you can't.

Then I think I would just go with my intuition because that way things are less complicated and also, I think I am going much beyond what my course material asks for.

Most people use intuition heavily. Just keep in mind that in a more advanced course, you'll have to revise your intuition.

 

[andyrk]

It's possible your material doesn't cover it, but most modern U.S. textbooks would at least have a statement in the "fine print" about when you can apply that procedure and when you can't.

Most people use intuition heavily. Just keep in mind that in a more advanced course, you'll have to revise your intuition.

What about the limit lim_{x\rightarrow a} e^{f(x)}? Can it be converted to e^{ lim_{x\rightarrow a} f(x)} without having the knowledge of composite limit theorem? Does this conversion involve the theorem or can it be simply converted without any hassles?

 

[andyrk]

You can think of it as a bet you make with a friend: Your friend has a function f (f because it is his favorite function) and some point c. Now the bet goes like this: Your friend bets to any number $\epsilon>0$ you tell him he can give you some $\delta>0$ such that for all points x' which are not farer from c than $\delta$ (which means $|x'-c| < \delta$) the function values f(x') and f(c) don't differ by more than $\varepsilon$ (this means $|f(x')-f(c)| < \varepsilon$).

So, if you find any point $x'$ for which $|x'-c|<\delta$ AND $|f(x')-f(c)|>\varepsilon$ your friend has failed to deliver a fitting $\delta > 0$ and you win the bet. You can try as often as you like with any $\varepsilon>0$ you want, your friend always has to give you a fitting $\delta>0$, and it is enough for you to find a single $x'$ with $|f(x')-f(c)|>\varepsilon$ for you to win. So you are in quite a good position.

However, if you can never find such an $\varepsilon$ and $x'$ which breaks the deal, your friend wins, shouts out the battlecry 'f IS CONTINUOUS AT c!' and pours an icebucket over your head. If on the other hand you find an $\varepsilon$ with a fitting $x'$, you may shout 'f IS NOT CONTINUOUS AT c!' victoriously and push him into a bathtub of vanilla pudding. So it's always quite a mess, but also a good day for mathematics.

Okay. This is out of my reach too. I think it would be better for me if I don't waste my time understanding this proof because no matter how hard I try I can't make any sense of it. I think my course material doesn't require to study such proofs and hence I should save the effort and utilize it somewhere which would yield me results. Its simply a waste of time studying things which I am not supposed to. :O

 

[Stephen Tashi]

What about the limit lim_{x\rightarrow a} e^{f(x)}? Can it be converted to e^{ lim_{x\rightarrow a} f(x)} without having the knowledge of composite limit theorem? Does this conversion involve the theorem or can it be simply converted without any hassles?

The "conversion" uses composite limit theorem. It uses the fact that e^x is continuous at each real number x = L. So if \lim_{x \rightarrow a} g(x) = L then e^x is continuous at x= L.

 

[andyrk]

The "conversion" uses composite limit theorem. It uses the fact that e^x is continuous at each real number x = L. So if \lim_{x \rightarrow a} g(x) = L then e^x is continuous at x= L.

Okay. But I would assume that it doesn't since it makes my life much tougher than if it hadn't. oo)

 

[Stephen Tashi]

Okay. But I would assume that it doesn't since it makes my life much tougher than if it hadn't. oo)

To cope with life, perhaps it is necessary to assume certain falsehoods. :)

 

[DarthMatter]

Okay. This is out of my reach too. I think it would be better for me if I don't waste my time understanding this proof because no matter how hard I try I can't make any sense of it. I think my course material doesn't require to study such proofs and hence I should save the effort and utilize it somewhere which would yield me results. Its simply a waste of time studying things which I am not supposed to. :O

I'm not giving up yet. And neither should you. :) You could also try this applet: https://www.geogebratube.org/student/m5258 . Try playing around with $\delta$, $\varepsilon$ and $L$ a little. If it all fits together, according to the definition, the correctly covered region of the function turns green.

 

[andyrk]

I'm not giving up yet. And neither should you. :) You could also try this applet: https://www.geogebratube.org/student/m5258 . Try playing around with $\delta$, $\varepsilon$ and $L$ a little. If it all fits together, according to the definition, the correctly covered region of the function turns green.

I think its best for me to give up right now. Because otherwise I would keep stressing too much about it and I wouldn't be able to complete the syllabus (which I presume is more important than studying something which is not required to be studied) :D. Perhaps, some other time...It would be a lot easier for me if I don't study this proof at all and treat it with indifference as if it never existed. My life's much easier that way. :) I might not be clear with the proof, but this is not the right time I think.

 

[DarthMatter]

Ok, so don't 'study' it. Play around with the applet a little until your have had enough and then take a nap or walk the dog. Jesus, people take mathematics so seriously these days.

 

[andyrk]

Hahah! :D

 

[Fredrik]

Okay. This is confusing. In the expression lim_{x \rightarrow a} h(g(x)) the argument of h() will become exactly L not near L. Because whenever we evaluate a limit we say "this limit is equal to" and not "this limit is near to". So similarly, the argument of h() would be equal to L not near L. If you think I didn't understand what you are trying to say, please explain it further to me.

It's appropriate to refer to the $\lim_{x\to c}$ as "the limit of $f$ at $c$", or as "the limit of $f(x)$ as $x$ goes to $c$". But consider the $f$ defined by
$$f(x)=\begin{cases}
1 & \text{if }x=c\\
0 & \text{if }x\neq c
\end{cases}$$ for all real numbers $x$. We have $\lim_{x\to c}f(x)=0\neq 1=f(c)$. This reflects the fact that the limit $\lim_{x\to c}f(x)$ is determined by the values of $f$ near $c$, not by the value of $f$ at $c$ (i.e. the number $f(c)$). The latter doesn't affect the limit at all.

Isn't it obvious that lim_{x \rightarrow a} h(g(x)) leads to h(L) because lim_{x \rightarrow a} g(x) = L??

No. The statement is true when $h$ is continuous at $L$, but consider $g(x)=Lx/a$ and
$$h(x)=\begin{cases}
1 & \text{if }x=L\\
0 & \text{if }x\neq L
\end{cases}.$$ We have $\lim_{x\to a}g(x)=g(a)=La/a=L$, but we also have $h(g(x))=0$ for all real numbers $x$ other than $a$, and this implies that
$$\lim_{x\to a}h(g(x))=0\neq 1=h(L)=h\left(\lim_{x\to a} g(x)\right).$$

I could never make any sense of the ε-δ proof. And I don't think I will be able to until some miracle happens. :(

You just need to study the definition of "limit". It's not that hard. OK, it's hard, but not so hard that you need a miracle.

Once you understand the definition and have done a few basic exercises about the definition of limit, I think you will find the proof I posted fairly easy to follow.

My course material says nothing about composite limit theorem. It says to substitute only in evaluating simple limits not composite. And infact I never encountered anywhere in all my problem sets or course material any reference to composite limits. I just encountered- lim_{x\rightarrow a} e^{f(x)} being equated to e^{lim_{x\rightarrow a} f(x)}. Does this require the understanding of composite function limit theorem? My course material is just advanced high school calculus (maybe that's the reason it doesn't cover it).

Since the exponential function is continuous, the theorem I proved ensures that those two are equal.

 

[andyrk]

It's appropriate to refer to the $\lim_{x\to c}$ as "the limit of $f$ at $c$", or as "the limit of $f(x)$ as $x$ goes to $c$". But consider the $f$ defined by
$$f(x)=\begin{cases}
1 & \text{if }x=c\\
0 & \text{if }x\neq c
\end{cases}$$ for all real numbers $x$. We have $\lim_{x\to c}f(x)=0\neq 1=f(c)$. This reflects the fact that the limit $\lim_{x\to c}f(x)$ is determined by the values of $f$ near $c$, not by the value of $f$ at $c$ (i.e. the number $f(c)$). The latter doesn't affect the limit at all.No. The statement is true when $h$ is continuous at $L$, but consider $g(x)=Lx/a$ and
$$h(x)=\begin{cases}
1 & \text{if }x=L\\
0 & \text{if }x\neq L
\end{cases}.$$ We have $\lim_{x\to a}g(x)=g(a)=La/a=L$, but we also have $h(g(x))=0$ for all real numbers $x$ other than $a$, and this implies that
$$\lim_{x\to a}h(g(x))=0\neq 1=h(L)=h\left(\lim_{x\to a} g(x)\right).$$You just need to study the definition of "limit". It's not that hard. OK, it's hard, but not so hard that you need a miracle.

Once you understand the definition and have done a few basic exercises about the definition of limit, I think you will find the proof I posted fairly easy to follow.Since the exponential function is continuous, the theorem I proved ensures that those two are equal.

Why? I still don't understand that why should h(x) be continuous at x = L for the two expressions to be equal?

 

[Fredrik]

Why? I still don't understand that why should h(x) be continuous at x = L for the two expressions to be equal?

I just showed you (in post #66) an example where $\lim_{x\to a}h(g(x))\neq h(\lim_{x\to a}g(x))$. I have also proved (in post #38) that continuity implies equality.

Why do you think that $\lim_{x\to a}h(g(x))$ should be equal to $h(\lim_{x\to a}g(x))$ even when $g(a)\neq\lim_{x\to a} g(x)$?

 

[andyrk]

I just showed you (in post #66) an example where $\lim_{x\to a}h(g(x))\neq h(\lim_{x\to a}g(x))$. I have also proved (in post #38) that continuity implies equality.

Why do you think that $\lim_{x\to a}h(g(x))$ should be equal to $h(\lim_{x\to a}g(x))$ even when $g(a)\neq\lim_{x\to a} g(x)$?

Lets forget the proof at the moment. The thing is that I consider $\lim_{x\to a}h(g(x))$ = h(g(a)) = h(L). Because I just come straightaway to evaluating the limit inside h(). I do that because I find no reason to check whether h is continuous at x = L or not. So as I have no reason (but there exists some reason which I am not able to understand for this whole thread) I don't check whether h() is continuous at x = L or not. If I should, it would have come automatically, but when I try solving it, it just doesn't come like that. My intuition tends to evaluate the limit straightaway. Maybe because it still hasn't settled in my subconsciousness that why h() needs to be continuous at x = L?

 

[DarthMatter]

Why should h(x) be continuous at a' to evaluate $\lim_{x\rightarrow a'} h(x)$?

 

[andyrk]

Why should h(x) be continuous at a' to evaluate $\lim_{x\rightarrow a'} h(x)$?

Exactly, it need not be. What are you wanting to say?

 

[DarthMatter]

Ok. Let's say $h(x)=1/x$ for $x\neq 0$, $h(0)=0$ (by definition). h is not continuous at 0.

So, let's approach 0 from the right by putting $a_n=\frac{1}{n}$. We then know $lim_{x\rightarrow 0}\frac{1}{x}=lim_{n\rightarrow\infty}\frac{1}{\frac{1}{n}}=lim_{n\rightarrow\infty}n=\infty$.

Now let's say we want to approach from the left: $a_n=-\frac{1}{n}$. We then get $lim_{x\rightarrow 0}\frac{1}{x}=lim_{n\rightarrow\infty}\frac{1}{-\frac{1}{n}}=lim_{n\rightarrow\infty}-n=-\infty$.



So which one is the right $lim_{x\rightarrow 0} h(x)$ now?

 

[andyrk]

Ok. Let's say $h(x)=1/x$ for $x\neq 0$, $h(0)=0$ (by definition). h is not continuous at 0.

So, let's approach 0 from the right by putting $a_n=\frac{1}{n}$. We then know $lim_{x\rightarrow 0}\frac{1}{x}=lim_{n\rightarrow\infty}\frac{1}{\frac{1}{n}}=lim_{n\rightarrow\infty}n=\infty$.

Now let's say we want to approach from the left: $a_n=-\frac{1}{n}$. We then get $lim_{x\rightarrow 0}\frac{1}{x}=lim_{n\rightarrow\infty}\frac{1}{-\frac{1}{n}}=lim_{n\rightarrow\infty}-n=-\infty$.



So which one is the right $lim_{x\rightarrow 0} h(x)$ now?

None of them. Because such a limit does not exist at x = 0. But this does not mean that it is impossible for a limit to exist at a point at which the function is discontinuous. A limit can exist at a point even though the function is discontinuous at that point. I think I am right with this.

 

[DarthMatter]

None of them. Because such a limit does not exist at x = 0. But this does not mean that it is impossible for a limit to exist at a point at which the function is discontinuous. I think I am right with this.

Yes it is impossible! :) If you mean the $\lim_{x\rightarrow a'} h(x)$. Because you will (in simple cases) get different results for your $\lim_{x\rightarrow a'}h(x)$ if you come from the left or if you come from the right. In the not so simple cases it may even depend on how exactly you approach $a'$.

But at the end of the day 'the' $\lim_{x\rightarrow a'} h(x)$ ist only defined if h(x) is continuous at a', and not depending on how you get near to a'. Because otherwise a lot of many different values for $\lim_{x\rightarrow a'} h(x)$ may be candidates, and none is better than the other.

 

[andyrk]

Yes it is impossible! :) If you mean the $\lim_{x\rightarrow a'} h(x)$. Because you will (in simple cases) get different results for your $\lim_{x\rightarrow a'}h(x)$ if you come from the left or if you come from the right. In the not so simple cases it may even depend on how exactly you approach $a'$.

But at the end of the day 'the' $\lim_{x\rightarrow a'} h(x)$ ist only defined if h(x) is continuous at a', and not depending on how you get near to a'. Because otherwise a lot of many different values for $\lim_{x\rightarrow a'} h(x)$ may be candidates, and none is better than the other.

For a limit to exist, only LHL (Left Hand Limit) needs to be equal to RHL (Right Hand Limit). There are many cases in graphs/functions where at certain points LHL = RHL but they are not equal to f(a) ( meaning that the function is discontinuous at x = a). But the limit still exists.

 

[DarthMatter]

:w This is true. You are right. But if h(x) is continuous, you can be sure that the limit exists. If h(x) is not continuous a', you just don't know for sure. Continuous also means you can replace the limit at a' with the function value at a'.

 

[andyrk]

:w This is true. You are right. But if h(x) is continuous, you can be sure that the limit exists.

So why is it impossible for the limit to exist in such cases? And also, how does this resolve the concern for h(x) being continuous at x = L for the limit to be h(g(a)) = h(L)?? And why do I need to be sure that the limit exists?

 

[DarthMatter]

It is not impossible. I was wrong. The example you gave shows it. But you need to be sure the limit exists in order to talk about $\lim_{x\rightarrow a'}h(x)$ as a certain real number. The easiest way to ensure it is to know that h(x) is continuous at a'.

The second thing is, you want to replace $lim_{x\rightarrow a'} h(x)=h(a')$, which is always wrong for discontinuous functions.

 

[andyrk]

It is not impossible. I was wrong. The example you gave shows it. But you need to be sure the limit exists in order to talk about $\lim_{x\rightarrow a'}h(x)$ as a certain real number. The easiest way to ensure it is to know that h(x) is continuous at a'.

The second thing is, you want to replace $lim_{x\rightarrow a'} h(x)=h(a')$, which is always wrong for discontinuous functions.

But what does this have to do with $lim_{x\rightarrow a} h(g(x))$ ?? Why do we need to ensure that h(x) should be continuous as x = L? Instead, shouldn't we ensure that g(x) is continuous at x = a? And infact, why do we need to ensure even that? We simply need $lim_{x\rightarrow a} h]g(x)=L$ (say). Once we evaluate this, then the expression $lim_{x\rightarrow a'} h(g(x))$ is easily evaluated and comes out to be h(L). What's the problem in this?

 

[DarthMatter]

The problem is, loosely speaking, that $g(x)$ can approach L from the left or from the right (similiar as $\frac{1}{n}$ and $-\frac{1}{n}$), and you have to be sure that both approaches yield the same value for $\lim_{x\rightarrow a}h(g(x))$.

Furthermore it has to be equal to h(L), which does not hold at a discontinuity .

 

[andyrk]

The problem is, loosely speaking, that $g(x)$ can approach L from the left or from the right (similiar as $\frac{1}{n}$ and $-\frac{1}{n}$), and you have to be sure that both approaches yield the same value for $\lim_{x\rightarrow a}h(g(x))$.

Furthermore it has to be equal to h(L), which does not hold at a discontinuity .

Why does it not hold at a discontinuity? $\lim_{x\rightarrow a}g(x) = L.$ and it need not be equal to g(a). So it still holds at discontinuity. So, afterwards, $\lim_{x\rightarrow a}h(g(x)) = h(L)$. So we talk of discontinuity while evaluating the limit on g(x) not on h(x). So why is everyone still saying that h(x) needs to be continuous at x = L and not g(x) should be continuous at x = a (even though it is not a requirement for the limit to exist)?

 

[DarthMatter]

Why does it not hold at a discontinuity? $\lim_{x\rightarrow a}g(x) = L.$ and it need not be equal to g(a). So it still holds at discontinuity. So, afterwards, $\lim_{x\rightarrow a}h(g(x)) = h(L)$. So we talk of discontinuity while evaluating the limit on g(x) not on h(x). So why is everyone still saying that h(x) needs to be continuous at x = L and not g(x) should be continuous at x = a (even though it is not a requirement for the limit to exist)?

It is because the expression $\lim_{x\rightarrow a}h(g(x))$ does not mean (by convention) 'put the limit of g into h'. It means 'put the function g into h and see what the limit is'. So in this expression, no limit with respect to g(x) is taken. The only limit taken is that of h(g(x)).

 

[andyrk]

It is because the expression $\lim_{x\rightarrow a}h(g(x))$ does not mean (by convention) 'put the limit of g into h'. It means 'put the function g into h and see what the limit is'. So in this expression, no limit with respect to g(x) is taken. The only limit taken is that of h(g(x)).

And how would you evaluate the limit on h(g(x))?

 

[DarthMatter]

In general, you just take the limit of $f(x)=h(g(x))$ as of any other function after you plugged g(x) into h. It can be very hard in general. But your theorem says: Under certain assumptions you can do other things which are simpler.

 

[andyrk]

In general, you just take the limit of $f(x)=h(g(x))$ as of any other function after you plugged g(x) into h. It can be very hard in general. But your theorem says: Under certain assumptions you can do other things which are simpler.

What are those assumptions and simpler things?

 

[DarthMatter]

What are those assumptions and simpler things?

Not sure what theorem you looked at in the first place. You should find the assumptions there. :)

Take this theorem for example:

https://www.proofwiki.org/wiki/Limit_of_Composite_Function

One possible assumption is that their f(x) (our h(x)) is continuous and that the proper limits of both functions exist. The consequences or simpler things are that, for our functions, $\lim_{x\rightarrow a}h(g(x))=h(\lim_{x\rightarrow a}g(x))$. You can also prove the consequences under other assumptions maybe, but the theorem never says there cannot be other cases. It just says, when these assumptions are met, this is enough, so you can do the simpler things.

 

[Fredrik]

Lets forget the proof at the moment. The thing is that I consider $\lim_{x\to a}h(g(x))$ = h(g(a)) = h(L). Because I just come straightaway to evaluating the limit inside h(). I do that because I find no reason to check whether h is continuous at x = L or not. So as I have no reason (but there exists some reason which I am not able to understand for this whole thread) I don't check whether h() is continuous at x = L or not. If I should, it would have come automatically, but when I try solving it, it just doesn't come like that. My intuition tends to evaluate the limit straightaway. Maybe because it still hasn't settled in my subconsciousness that why h() needs to be continuous at x = L?

The fact that you think of $\lim_{x\to a}h(g(x))$ as equal to $h(g(a))$ suggests that you still think that the former is somehow determined by the number $a$ rather than by how the function behaves on a set that doesn't even include $a$. Please look at my post #66 again, in particular the part that replies to the first quote in that post.

 

[andyrk]

Not sure what theorem you looked at in the first place. You should find the assumptions there. :)

Take this theorem for example:

https://www.proofwiki.org/wiki/Limit_of_Composite_Function

One possible assumption is that their f(x) (our h(x)) is continuous and that the proper limits of both functions exist. The consequences or simpler things are that, for our functions, $\lim_{x\rightarrow a}h(g(x))=h(\lim_{x\rightarrow a}g(x))$. You can also prove the consequences under other assumptions maybe, but the theorem never says there cannot be other cases. It just says, when these assumptions are met, this is enough, so you can do the simpler things.

The whole theorem says nothing about why we can't $\lim_{x\rightarrow a}h(g(x))=h(\lim_{x\rightarrow a}g(x))$ straightaway?

 

[Fredrik]

The whole theorem says nothing about why we can't $\lim_{x\rightarrow a}h(g(x))=h(\lim_{x\rightarrow a}g(x))$ straightaway?

Right, the theorem tells you that this equality holds when h is continuous, and doesn't say anything about whether the equality holds when h is not continous. The proof tells you how to use the continuity of h to see that the equality holds when h is continuous. What tells you that the equality doesn't hold in general are the counterexamples. I showed you one of the simplest possible counterexamples in (the middle part of) post #66.

 

[andyrk]

This is post #66.

It's appropriate to refer to the $\lim_{x\to c}$ as "the limit of $f$ at $c$", or as "the limit of $f(x)$ as $x$ goes to $c$". But consider the $f$ defined by
$$f(x)=\begin{cases}
1 & \text{if }x=c\\
0 & \text{if }x\neq c
\end{cases}$$ for all real numbers $x$. We have $\lim_{x\to c}f(x)=0\neq 1=f(c)$. This reflects the fact that the limit $\lim_{x\to c}f(x)$ is determined by the values of $f$ near $c$, not by the value of $f$ at $c$ (i.e. the number $f(c)$). The latter doesn't affect the limit at all.No. The statement is true when $h$ is continuous at $L$, but consider $g(x)=Lx/a$ and
$$h(x)=\begin{cases}
1 & \text{if }x=L\\
0 & \text{if }x\neq L
\end{cases}.$$ We have $\lim_{x\to a}g(x)=g(a)=La/a=L$, but we also have $h(g(x))=0$ for all real numbers $x$ other than $a$, and this implies that
$$\lim_{x\to a}h(g(x))=0\neq 1=h(L)=h\left(\lim_{x\to a} g(x)\right).$$You just need to study the definition of "limit". It's not that hard. OK, it's hard, but not so hard that you need a miracle.

Once you understand the definition and have done a few basic exercises about the definition of limit, I think you will find the proof I posted fairly easy to follow.Since the exponential function is continuous, the theorem I proved ensures that those two are equal.

I think I get the general idea in this post. It is that $$\lim_{x\to a}h(g(x))$$ = h(g(a number near a)) = h(a number near L) = 0.

And $$h\left(\lim_{x\to a} g(x)\right)$$ = h(g(exactly a)) = h(exactly L) = 1. But why does this happen? Is it because in the first part, there is no limit on g(x) as in the second part? But my question is that why is there no limit on g(x) even though there should be in the first part?

 

[DarthMatter]

Why should there be a limit on g(x) in the first part? That would mean you are not talking about the limit $\lim_{x \rightarrow a} f(x)$ of the function f(x) for $f(x)=h(g(x))$, but about some other expression where a limit of g(x) is directly involved. You just have to check the definition to see that there is no reason to assume that by definition $f(x)$ is different in respect to the evaluation of limits because it is a composite function.

Is it because in the first part, there is no limit on g(x) as in the second part?

Yes. It shows that, in general and without further assumptions, $h(\lim_{x\rightarrow a} g(x))\neq \lim_{x\rightarrow a} h(g(x))$.

 

[Fredrik]

I think I get the general idea in this post. It is that $$\lim_{x\to a}h(g(x))$$ = h(g(a number near a)) = h(a number near L) = 0.

With this choice of $h$ and $g$, I suppose this is accurate enough. But in general (when $h$ and $g$ aren't assumed to be this nice), it's possible that there's no number $b$ such that $\lim_{x\to a} h(g(x))=h(g(b))$. Consider e.g. the $f$ defined by
$$f(x)=\begin{cases}-1 & \text{if }x=0\\ x^2 & \text{if }x\neq 0\end{cases}$$ for all real numbers $x$. We have $\lim_{x\to 0}f(x)=0$, but $f(0)=-1$ and $f(x)> 0$ for all $x$ in the domain of $f$ except $0$.

And $$h\left(\lim_{x\to a} g(x)\right)$$ = h(g(exactly a)) = h(exactly L) = 1. But why does this happen? Is it because in the first part, there is no limit on g(x) as in the second part? But my question is that why is there no limit on g(x) even though there should be in the first part?

I don't quite understand the question. Are you asking why $\lim_{x\to a}g(x)$ is equal to $g(a)$? It's because the specific $g$ I chose is continuous at $a$. Its graph is just a straight line through the origin, with slope $L/a$. The equality $\lim_{x\to a}g(x)=g(a)$ doesn't hold for a $g$ that isn't continuous at $a$. That can be taken as the definition of "continous at $a$".

 

[andyrk]

With this choice of $h$ and $g$, I suppose this is accurate enough. But in general (when $h$ and $g$ aren't assumed to be this nice), it's possible that there's no number $b$ such that $\lim_{x\to a} h(g(x))=h(g(b))$. Consider e.g. the $f$ defined by
$$f(x)=\begin{cases}-1 & \text{if }x=0\\ x^2 & \text{if }x\neq 0\end{cases}$$ for all real numbers $x$. We have $\lim_{x\to 0}f(x)=0$, but $f(0)=-1$ and $f(x)> 0$ for all $x$ in the domain of $f$ except $0$.I don't quite understand the question. Are you asking why $\lim_{x\to a}g(x)$ is equal to $g(a)$? It's because the specific $g$ I chose is continuous at $a$. Its graph is just a straight line through the origin, with slope $L/a$. The equality $\lim_{x\to a}g(x)=g(a)$ doesn't hold for a $g$ that isn't continuous at $a$. That can be taken as the definition of "continous at $a$".

No. I am not asking that. What I am asking is that when we write the limit - $\lim_{x\to a} h(g(x))$, the general intuition is (in order to get to your explanation) as x tends towards a, g(x) tends towards L (that does not mean that g(x) actually ever reaches L). That's because that is how a limit is defined, i.e. as x approaches a number, f(x) approaches some other number. And that number is equated to be the limit of f(x) when we operate the limit on f(x). What I am asking is that while writing the limit, $\lim_{x\to a} h(g(x))$ why don't we apply the limit to g(x) so as to clearly say that as x approaches a g(x) approaches L (and then we can replace the limit $\lim_{x\to a} h(g(x))$ by h(L) directly) ? Is it because of the way a limit is defined? The definition of a limit is "In mathematics, a limit is the value that a function or sequence "approaches" as the input or index approaches some value". Is that the reason why we don't take the limit on g(x) in $\lim_{x\to a} h(g(x))$? Because we are concerned about the value that h(x) approaches to because that is the function that we are talking about? Is that right? g(x) just acts as an input which approaches a value L when x approaches a. So, for h(x), the x(which is g(x) in our case) in h(x) approaches L and then we have to see what h(x) approaches to when x approaches L? So does the limit essentially evaluate to $\lim_{x\to L} h(x)$ on simplification?

 

[Fredrik]

No. I am not asking that. What I am asking is that when we write the limit - $\lim_{x\to a} h(g(x))$, the general intuition is (in order to get to your explanation) as x tends towards a, g(x) tends towards L (that does not mean that g(x) actually ever reaches L). That's because that is how a limit is defined, i.e. as x approaches a number, f(x) approaches some other number. And that number is equated to be the limit of f(x) when we operate the limit on f(x). What I am asking is that while writing the limit, $\lim_{x\to a} h(g(x))$ why don't we apply the limit to g(x) so as to clearly say that as x approaches a g(x) approaches L (and then we can replace the limit $\lim_{x\to a} h(g(x))$ by h(L) directly) ?

We don't because my counterexample and others show that this leads to nonsense results like 0=1. As I said, the value of the expression $\lim_{x\to a}h(g(x))$ (i.e. what number this string of text represents) isn't determined by the value of $h(g(a))$, which may not even be defined, but by the behavior of $h\circ g$ (the function that most people refer to as $h(g(x))$) on a set that doesn't include $a$. I'm not sure what else I can tell you.

Is it because of the way a limit is defined? The definition of a limit is "In mathematics, a limit is the value that a function or sequence "approaches" as the input or index approaches some value".

That's not a definition. It's a suggestion about how to think about limits. It can also be viewed as the motivation for the actual definition, which goes like this:

A real number $L$ is said to be a limit at $a$ of the function $f$, if for all $\varepsilon>0$, there's a $\delta>0$ such that the following implication holds for all $x$ in the domain of $f$
$$0<|x-a|<\delta\ \Rightarrow\ |f(x)-L|<\varepsilon.$$ This means that if we plot the graph, and I draw two horizontal lines at the same distance (we call this distance $\varepsilon$) from $L$ on the $y$ axis, then regardless of how close to $L$ I drew them, you can draw two vertical lines at the same distance (we call this distance $\delta$) from $a$ on the $x$ axis, such that except for the single point $(a,L)$, the part of the graph that's between your two vertical lines is also between my two horizontal lines.

As DarthMatter told you earlier, you can think of this as a game that we're playing. To say that $L$ is a limit of $f$ at $a$ is to say that you can always win the game by drawing your vertical lines close enough to $a$.

People often simplify the explanation to "you can make $f(x)$ arbitrarily close to $L$ by choosing $x$ close enough to $a$". This captures part of the idea, but is inadequate as a definition.

 

[andyrk]

What about the rest of the explanation I gave? Is that correct?

 

[DarthMatter]

No. I am not asking that. What I am asking is that when we write the limit - $\lim_{x\to a} h(g(x))$, the general intuition is (in order to get to your explanation) as x tends towards a, g(x) tends towards L (that does not mean that g(x) actually ever reaches L). That's because that is how a limit is defined, i.e. as x approaches a number, f(x) approaches some other number. And that number is equated to be the limit of f(x) when we operate the limit on f(x).

I think that is good picture to get some intuition for the concept.

What I am asking is that while writing the limit, $\lim_{x\to a} h(g(x))$ why don't we apply the limit to g(x) so as to clearly say that as x approaches a g(x) approaches L (and then we can replace the limit $\lim_{x\to a} h(g(x))$ by h(L) directly) ? Is it because of the way a limit is defined? The definition of a limit is "In mathematics, a limit is the value that a function or sequence "approaches" as the input or index approaches some value". Is that the reason why we don't take the limit on g(x) in $\lim_{x\to a} h(g(x))$? Because we are concerned about the value that h(x) approaches to because that is the function that we are talking about? Is that right? g(x) just acts as an input which approaches a value L when x approaches a. So, for h(x), the x(which is g(x) in our case) in h(x) approaches L and then we have to see what h(x) approaches to when x approaches L?

Maybe it helps to remember that (as you already wrote) under the assumption that all our limits exist $x\rightarrow a$ implies $g(x) \rightarrow L$. So in general under these asumption calculating the limit $\lim_{x\rightarrow a} f(g(x))$ is just another way of writing $\lim_{x\rightarrow L} h(x)$. However, that limit does not have to be equal to $h(L)$.
Oh, you already saw that:

So does the limit essentially evaluate to $\lim_{x\to L} h(x)$ on simplification?

Good job! I think this is (under the assumption that the limits exist) a key idea. :)

 

[DarthMatter]

PS: If you want to check out the $\varepsilon$-$\delta$-stuff again with some certain numerical values for $\epsilon$ and $\delta$, I think 'Elementary Analysis: The Theory of Calculus' by Keneth A. Ross, §7 does a good job on that.§7 is on sequences, but that is a foundation to understand it in the area of functions.

 

[Fredrik]

What about the rest of the explanation I gave? Is that correct?

No comment about the things I did explain? You seem to often focus on the wrong things.

I'm not sure I followed your argument, but if you're asking if $\lim_{x\to a}g(x)=L$ implies that $\lim_{x\to a}h(g(x))=\lim_{x\to L} h(x)$ regardless of whether $h$ is continuous, then the answer is that it depends on $g$. This equality holds when $h$ and $g$ are the functions in my example, even though $h$ isn't continuous, but I can and will show you a $g$ such that $\lim_{x\to a}g(x)=L$ but $\lim_{x\to a}h(g(x))\neq \lim_{x\to L} h(x)$.

If $h$ and $g$ are defined as in my example (see the middle part of post #66), then we have
$$\lim_{x\to a}\underbrace{h(g(x))}_{=0\text{ when }x\neq a} =0 =\lim_{x\to L}\underbrace{h(x)}_{=0\text{ when }x\neq L},$$ so the equality holds, but note that we also have
$$h\Big(\lim_{x\to a}g(x)\Big) = h(L)= h\left(\lim_{x\to L} x\right).$$ Since $h(L)=1$, these results mean that $\lim_{x\to a}h(g(x))\neq h(L)$ och $\lim_{x\to L}h(x)\neq h(L)$.

Now suppose that we instead define $g$ by
$$g(x)=\begin{cases}0 & \text{when }x=a\\ L & \text{when }x\neq a\end{cases}$$ for all real numbers $x$. We still have $\lim_{x\to a}g(x)=L$, but now we have
$$\lim_{x\to a}\underbrace{h(g(x))}_{=1\text{ when }x\neq a} =1$$ and
$$\lim_{x\to L}\underbrace{h(x)}_{=0\text{ when }x\neq L} =0.$$ So
$$\lim_{x\to a}h(g(x))=1\neq 0= \lim_{x\to L} h(x).$$

 

[DarthMatter]

Good point, Fredrik. So I have to take that back. But I still think it is a good idea andyrk had.
Do you think it would be sufficient to further assume $g(x)\neq L$ for all x to get $\lim_{x \rightarrow a} h(g(x)) = \lim_{x \rightarrow L} h(x)$ ?

 

[Fredrik]

Do you think it would be sufficient to further assume $g(x)\neq L$ for all x to get $\lim_{x \rightarrow a} h(g(x)) = \lim_{x \rightarrow L} h(x)$ ?

Yes, that's sufficient, when we're dealing with this particular $h$.