Understanding limit of exponents

[andyrk]

The composite limit theorem says that if certain conditions are met then the two expressions are equal. In the example of $lim_{x \rightarrow 5}h(g(x))$ the conditions are not met. So you cannot conclude the two expressions are equal.

You seem to think there is a rule that $\lim_{x\rightarrow a} h(g(x))$ can always be "rewritten" as $h(\lim_{x\rightarrow a} g(x))$ regardless of what properties $h$ has. That is a false belief.
The $\lim_{x\rightarrow a} h(g(x))$ does depend on the values of $x$ in the vicinity of $L$.
Think about $lim_{x \rightarrow L} h(x)$.

The conditions are (as the video states)-

" If $f$ and $g$ are functions such that-

$lim_{x \rightarrow c} g(x) = L$

and

$lim_{x \rightarrow L} f(x) = f(L)$

then, $lim_{x \rightarrow c} f(g(x)) = f(lim_{x \rightarrow c} g(x)) = f(L)$ "

So how can you say that the conditions are not met? According to what is written above, they do.

In general, you should also know that there are examples where $lim_{x \rightarrow a} f(x) = L$ and $f(a)$ is not equal to $L {/itex] .$

[itex] <br /> This is possible only if $f(x)$ is not continuous at $x = a$. Right? Because for a limit to exist, the function need not be continuous at the point where the limit is being evaluated. Am I right?[/itex]

 

[Fredrik]

The conditions are (as the video states)-

" If $f$ and $g$ are functions such that-

$lim_{x \rightarrow c} g(x) = L$

and

$lim_{x \rightarrow L} f(x) = f(L)$

then, $lim_{x \rightarrow c} f(g(x)) = f(lim_{x \rightarrow c} g(x)) = f(L)$ "

So how can you say that the conditions are not met? According to what is written above, they do.

In the example with h(x)=0 for x<0, h(x)=2+x for x≥0 and g(x)=x-5 for all x, only the first of these conditions is met. We have $\lim_{x\to 5} g(x)=g(5)=0$, but the limit $\lim_{x\to 0} h(x)$ doesn't exist.

The second condition in the quote above can be thought of as the definition of what it means to say that f is continuous at L.

This is possible only if $f(x)$ is not continuous at $x = a$. Right? Because for a limit to exist, the function need not be continuous at the point where the limit is being evaluated. Am I right?

Yes.

 

[Fredrik]

Consider this simple example. Let $f$ be the function defined by
$$f(x)=\begin{cases}
1 & \text{if }x=0\\
0 & \text{if }x\neq 0
\end{cases}.$$ We have $f(\lim_{x\to 0} x)=f(0)=1\neq 0=\lim_{x\to 0}f(x)$. Clearly the probolem is that $f(0)$ isn't equal to $\lim_{x\to 0}f(x)$, i.e. that $f$ isn't continuous at $0$.

 

[andyrk]

In the example with h(x)=0 for x<0, h(x)=2+x for x≥0 and g(x)=x-5 for all x, only the first of these conditions is met. We have $\lim_{x\to 5} g(x)=g(5)=0$, but the limit $\lim_{x\to 0} h(x)$ doesn't exist.

The second condition in the quote above can be thought of as the definition of what it means to say that f is continuous at L.Yes.

Could you explain me the proof for this theorem? I have it but I don't understand it well enough-

"For a given $ε > 0$, find $∂ > 0$ such that-
$| f(g(x)) - f(L) | < ε$
Whenever $0 < |x - c| < ∂$"

Now as to how to find such $ε$ such that $ε > 0$-

$| f(u) - f(L) | < ε$ Whenever | μ - L | < ∂₁ (Didn't understand this at all)

| g(x) - f(L) | < ∂₁ , whenever $0 < |x - c| < ∂$ (didn't understand it again)

Let $μ = g(x)$ , $| f(g(x)) - f(L) | < ε$, whenever $0 < |x - c| < ∂$. (didn't understand this either)

 

[andyrk]

We have $f(\lim_{x\to 0} x)=f(0)=1\neq 0=\lim_{x\to 0}f(x)$. Clearly the problem is that $f(0)$ isn't equal to $\lim_{x\to 0}f(x)$

Okay, but first you said that $f(0)$ is equal to $\lim_{x\to 0}f(x)$. Then you say that it isn't. But I think it is. Which one am I supposed to believe anyways?

And anyhow, this example didn't make me understand the proof of the theorem even though it was really good and I appreciate you for that. :)

 

[Fredrik]

Okay, but first you said that $f(0)$ is equal to $\lim_{x\to 0}f(x)$. Then you say that it isn't.

No, I didn't. I said that $f(0)$ is equal to $f(\lim_{x\to 0}x)$ (you know that $\lim_{x\to 0}x=0$, right?) and then I said that it's not equal to $\lim_{x\to 0}f(x)$. Since $f(x)=0$ for all $x$ other than $0$, we have $\lim_{x\to 0}f(x)=\lim_{x\to 0}0 =0\neq 1=f(0)=f(\lim_{x\to 0} x)$.

But I think it is.

So which of the equalities I wrote down is wrong?

 

[DarthMatter]

Could you explain me the proof for this theorem? I have it but I don't understand it well enough-

"For a given $ε > 0$, find $∂ > 0$ such that-
$| f(g(x)) - f(L) | < ε$
Whenever $0 < |x - c| < ∂$"Now as to how to find such $ε$ such that $ε > 0$-

$| f(u) - f(L) | < ε$ Whenever | μ - L | < ∂₁ (Didn't understand this at all)

Here you are using that f is continuous at L (this means that the limit exists). In other words, if you move your $\mu$ in $f(\mu)$ 'near enough' to L, you can make $|f(\mu)-f(L)|$ as small as you like. (Formally: Smaller than any $\varepsilon > 0$.) Formally you can express moving $\mu$ very near to L by $|\mu-L|<\delta_1$. You should just think of all $\delta s$ (and $\varepsilon s$) as very small, arbitrarily small numbers bigger than zero. You choose your $\delta$s depending on how small you want to make the differences such as $|f(\mu)-f(L)|$. (Formally: You choose the current $\delta$ depending on its $\varepsilon$.)

To be honest I am not sure the rest of the proof is correct. I will try to fix it (or maybe make it easier to understand).

Lets use that the limit $\lim_{x\rightarrow c} g(x) = L$. This means we can make $|g(x)-L|$ smaller than any arbitrarily small number bigger than zero we can imagine, right? So let's make $|g(x)-L|$ smaller than the $\delta_1$ we used for $|\mu-L| < \delta_1$ before. We can always do this by putting x as near to c as is necessary, because the limit $\lim_{x\rightarrow c} g(x) = L$ exists. We are setting our epsilon to $\delta_1$ here, so to say. So if we choose |x-c| small enough, we will have $|g(x)-g(c)|=|g(x)-L| < \delta_1$.

Why did we choose $\delta_1$ as our epsilon for the second function g(x)? Because then we then can not only say something about g(x),but also about f(g(x)).

Here it comes:

$|g(x)-L| < \delta_1$ also means, as we found at the beginning of this post, $|f(g(x))-f(L)|=|f(g(x))-f(g(c))| < \varepsilon$. Therefore we have proven that we can make the difference between $f(g(x))$ and $f(L)$ as small as we want, by putting $x$ as near to c as necessary. Therefore $lim_{x\rightarrow c} f(g(x))=f(L) = f(\lim_{x\rightarrow c} g(x))$.

 

[Fredrik]

Here's my version of the definition, theorem and proof:

Definition: $f$ is said to be continuous at $L$ if $\lim_{x\to L}f(x)=f(L)$.

Theorem: If $\lim_{x\to c}g(x)=L$, and $f$ is continuous at $L$, then $\lim_{x\to c}f(g(x))=f(L)$.

Proof: Let $\varepsilon>0$ be arbitrary. Let $\delta_1$ be a positive real number such that the implication
$$0<|u-L|<\delta_1\ \Rightarrow\ |f(u)-f(L)|<\varepsilon$$ holds for all $u$ in the domain of $f$. (Such a $\delta_1$ exists because $f$ is continuous at $L$).

Let $\delta_2$ be a positive real number such that the implication
$$0<|x-c|<\delta_2\ \Rightarrow\ |g(x)-L|<\delta_1$$ holds for all $x$ in the domain of $g$. (Such a $\delta_2$ exists because $\lim_{x\to c}g(x)=L$).

Now let $x$ be an arbitrary real number such that the following statements are true:

1. $x$ is in the domain of $g$.
2. $g(x)$ is in the domain of $f$.
3. $0<|x-c|<\delta_2$.

By definition of $\delta_2$ we have $|g(x)-L|<\delta_1$.

We will prove that $|f(g(x))-f(L)|<\varepsilon.$ We will consider the possibilities $g(x)\neq L$ and $g(x)=L$ separately. First suppose that $g(x)\neq L$. Then we have $0<|g(x)-L|<\delta_1$. By definition of $\delta_1$, this implies that $|f(g(x))-f(L)|<\varepsilon$. Now suppose instead that $g(x)=L$. Then we have $f(g(x))=f(L)$ and therefore $|f(g(x))-f(L)|=0<\varepsilon.$ So regardless of whether $g(x)$ is equal to $L$, we have $|f(g(x))-L|<\varepsilon$. Since $x$ is an arbitrary real number that satisfies the three conditions above, this implies that $\lim_{x\to c}f(g(x))=L$.

 

[DarthMatter]

Maybe an example is in order to make the argument more clear. A very simple example is putting f(x)=x+2 and g(x)=x+1. You can make the difference between g(x') and g(0) arbitrarily small by letting $|x'-0| =|x'| < \delta_g$: In this case you get $|g(0)-g(x')| = |0+1-(x'+1)| = |x'|$. So if $|0-x'|<\delta_g$, also $|g(0)-g(x')| < \delta_g$. So for any given $\varepsilon_g>0$, you can just set $\delta_g=\varepsilon_g$. Similiarly $|f(1)-f(x')|<\delta_f$, for $|1-x'|<\delta_f$. So for any given $\varepsilon_f>0$ you can just set $\delta_f=\varepsilon_f$ as well. This a special case, normally $\varepsilon$ and $\delta$ will be different.

Now let's look at he composition. $f(g(x))=x+3=h(x)$. Let's say we want to make the difference |h(0)-h(x')| smaller than a given $\varepsilon_h>0$. Therefore, following the above argument, we first choose $|x'-0| < \delta_h=\varepsilon_h$. From this we observe $|g(0)-g(x')| < \varepsilon_h$. But this can be rewritten as $|1-g(x')|<\varepsilon_h$. Remembering that $|1-x'|<\delta_f$ implies $|f(1)-f(x')|<\delta_f$, $|1-g(x')|<\varepsilon_h$ implies $|f(1)-f(g(x'))| < \varepsilon_h$. So we have proven that we can make the difference $|f(1)-f(g(x'))|$ smaller than any given $\varepsilon_h>0$. This means, by definition, $\lim_{x\rightarrow 0} h(x) = f(1) = f(\lim_{x\rightarrow 0}g(x))$.

 

[andyrk]

The expression $lim_{x \rightarrow a} h(g(x))$ is a limit that depends on what happens then the argument of the function $h()$ is near the value $L$ . By contrast the expression $h( lim_{x \rightarrow a} g(x) )$ depends only on the value of $h$ when the argument is exactly equal to $L$ .

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.

 

[DarthMatter]

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" 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.

The limit just is the real number the function value approaches when you move it closer and closer to a. You have to prove that you can also get this number by putting L into h as an argument before you can do it. So for understanding what $lim_{x \rightarrow a} h(g(x))$ is you need just the definition. Recognizing it is equal to $h(L)$ requires some mathematical work.

 

[andyrk]

You have to prove that you can also get this number by putting L into h as an argument before you can do it

Before I can do what? L is the number which the function value approaches to as x moves closer and closer to a. Do I need to prove this?

So for understanding what $lim_{x \rightarrow a} h(g(x))$ is you need just the definition. Recognizing it is equal to $h(L)$ requires some mathematical work.

Why does it need some mathematical work? Isn't it obvious that $lim_{x \rightarrow a} h(g(x))$ leads to $h(L)$ because $lim_{x \rightarrow a} g(x) = L$??

 

[DarthMatter]

Before I can do what? L is the number which the function value approaches to as x moves closer and closer to a. Do I need to prove this?

Not you personally. I can't control this anyway. No, you don't, but someone has to prove it (as a general theorem in a textbook, for example). :)

Why does it need some mathematical work? Isn't it obvious that $lim_{x \rightarrow a} h(g(x))$ leads to $h(L)$ because $lim_{x \rightarrow a} g(x) = L$??

The idea is correct, but if no one had proven it, how could you be sure? My point is basically that it does not directly follow from the definition of the limit that it is equal to $h(L)$. There is a theorem/proof in between. So it is maybe not that confusing when you keep in mind that no one ever said it does not matter where if you put the lim inside the argument or not. So you are right (you said you found this confusing): $lim_{x \rightarrow a} h(g(x))$ and $h(lim_{x \rightarrow a} g(x))$ are two mathematically very different things. In the first case you look at the limit of h(g) at some point, in the second cast you put some certain number into h (the limit of g). But as a result of a theorem, these very different mathematical concepts can be equal.

 

[andyrk]

So you are right (you said you found this confusing): $lim_{x \rightarrow a} h(g(x))$ and $h(lim_{x \rightarrow a} g(x))$ are two mathematically very different things. In the first case you look at the limit of h at some point, in the second cast you put some certain number into h (the limit of g). But as a result of a theorem, these very different mathematical concepts can be equal.

Eureka! I got that correct! But for some reason, Stephen Tashi doesn't agree with this. He says the two expressions are quite different, even though they come out to be equal just as you said.

Anyways, where was I? First of all, why was I asking this? It was to prove my reasoning that $lim_{x \rightarrow a} h(g(x))$ is not h(near to L) but h(exactly L). This was what I was arguing about since the very beginning. And it seems I have proven it right, have I, finally?

 

[andyrk]

And when we are evaluating $lim_{x \rightarrow a} h(g(x))$ why should we check that h(x) is continuous at x = L at all? It just needs to be defined at x = L, that's it! If it is, then the answer is h(L) regardless of whether it is continuous at x = L or not. Because the limit has got nothing to do with whether h(x) is continuous at x = L or not. It is concerned with only g(x) being continuous at x = a (as a is what x approaches to in the limit). Am I correct?

 

[DarthMatter]

Eureka! I got that correct! But for some reason, Stephen Tashi doesn't agree with this. He says the two expressions are quite different, even though they come out to be equal just as you said.

Anyways, where was I? First of all, why was I asking this? It was to prove my reasoning that $lim_{x \rightarrow a} h(g(x))$ is not h(near to L) but h(exactly L). This was what I was arguing about since the very beginning. And it seems I have proven it right, have I, finally?

Well the mathematical concepts are different. $lim_{x \rightarrow a} h(g(x))$ still is the real number you approach with your composite function as you let x go to a (you don't have to ever reach a, just come closer and closer). So I would agree with Stephen here. The theorem now says you can also get this number under some assumptions in another way, by inserting L into h. So you have two different mathematical objects which have the same value. So both is true in this case: $lim_{x\rightarrow a} h(g(x))$ still is the number you approach with h(g(x)) as x goes to a, but it is also equal to $h(L)$.

 

[DarthMatter]

And when we are evaluating $lim_{x \rightarrow a} h(g(x))$ why should we check that h(x) is continuous at x = L at all? It just needs to be defined at x = L, that's it! If it is, then the answer is h(L) regardless of whether it is continuous at x = L or not. Because the limit has got nothing to do with whether h(x) is continuous at x = L or not. It is concerned with only g(x) being continuous at x = a (as a is what x approaches to in the limit). Am I correct?

No, if h(x) is not continuous the limit is not guaranteed to exist.

 

[andyrk]

No, if h(x) is not continuous the limit is not guaranteed to exist.

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.

 

[Stephen Tashi]

Okay. This is confusing. In the expression $lim_{x \rightarrow a} h(g(x))$ the argument of $h()$ will become exactly $L$ not near $L$.

No. it is in the expression $h(lim_{x \rightarrow a} g(x) )$ that the argument of $h()$ is a specific number (if the limit exists), because, as you say, a limit is a specific number. There is nothing in the definition of $\lim_{x\rightarrow a} h(g(x))$ that says $g(a)$ must equal $L = lim_{x \rightarrow a} g(x)$

One can get some understanding of limits using common speech (e.g. talking about things getting near to things etc.). However, to fully understand limits, you have to deal with the epsilon-delta definitions. It takes some time and experience to deal with the technical definitions and if you aren't ready to do that yet, then you must grope your way using intuition and imprecise language. You seem to have a wrong intuition that a limit is a number that "must actually be reached".

 

[andyrk]

There is nothing in the definition of $\lim_{x\rightarrow a} h(g(x))$ that says $g(a)$ must equal $L = lim_{x \rightarrow a} g(x)$

That's because my course material says to directly substitute the value (for such limits) to which x is approaching, in the function (g(x) in this case). So that is the reason I substituted a in $g(x)$ in $lim_{x \rightarrow a} g(x) = L$ giving me $L = g(a) = lim_{x \rightarrow a} g(x)$

You seem to have a wrong intuition that a limit is a number that "must actually be reached".

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 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. I am just concerned with what the limit evaluates to. I know that the function never reaches the value which the limit evaluated to but that is not what I am concerned about.

 

[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)