Understanding limit of exponents

[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 (similar 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 (similar 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?