# Understanding limit of exponents

1. Dec 24, 2014

### andyrk

This might be a pretty stupid question. But why is it that while applying limits to an exponential function like- $\lim_{x\rightarrow 0} e^{f(x)}$ we move the limit to only the part of the expression which involves the variable on which the limit is being evaluated and hence we now write it as- $e^{(\lim_{x\rightarrow 0} f(x))}$? Can't we evaluate the limit without reducing the terms inside the limit?

What I mean is earlier, the limit included $e$ withing itself too. But when we simplified it, the $e$ came out of the limit and instead the limit was being operated on only $f(x)$. Can we do this? Is there any formal rule or formula for this like other formulas for evaluating and simplifying limits?

So my question is basically that is this a rule or formula which we follow or do we do it just because of logic? (The logic of applying the limit only to the part of the expression which gets affected by the limit)

Last edited: Dec 24, 2014
2. Dec 24, 2014

### Staff: Mentor

I think its the logic of it. Basically you could look at the expression as a mapping of x to e^f(x) and so then you could look at the x to f(x) as x approaches 0.

If you think about it knowing what happens to f(x) in the limit i.e. if it has a value means you can plug it into the e^(limit value) to get the limit value of the original expression.

3. Dec 24, 2014

### andyrk

What did you exactly mean by this statement? I didn't understand you fully. Earlier we were looking at x maps to e^f(x). Then how can we look at x maps to f(x) afterwards? Shouldn't it be x maps to e^f(x) all throughout?

4. Dec 24, 2014

### DarthMatter

You can do this if $g(x)$ is an continuous function.

Let g(x) be a continuous function such that $\lim_{x\rightarrow x_0}g(x)=a$. Then for each $\epsilon > 0$ there is a ball $B_\epsilon$ around $x_0$ such that $|g(x)-a| < \epsilon$ for all $x \in B_\epsilon$.
Also for all $x$ $|\exp(g(x))-\exp(a)|=|\exp(g(x)-a)-1|\cdot \exp(+a)$. Now for each $\epsilon' > 0$ choose $B_{\epsilon'}$ such that $|g(x)-a| < \ln(1+\frac{\epsilon'}{\exp(a)})$. Using this in the expression above and the positive derivative of the e-function yields $|\exp(g(x))-\exp(a)| < \epsilon'$. It follows that $\lim_{x\rightarrow x_0}\exp(g(x)) = \exp(g(a))$.

5. Dec 24, 2014

### DarthMatter

How can I get the arrows under the $\lim$?

6. Dec 24, 2014

### Staff: Mentor

I was thinking of the calculus situation where you y=f(g(x)) and you want to find the limit when x goes to zero then you functionally decompose it by looking at the limit for g(x) and finding that plug it into f( ) to get the final answer.

7. Dec 24, 2014

### Stephen Tashi

As DarthMatter said, if $h(x)$ is continuous at $x = b$ and $lim_{x \rightarrow a} f(x) = b$ you can assert that $lim_{x\rightarrow a} h(f(x)) = h ( lim_{x \rightarrow a} f(x) )$.

In you example we have the special case $a = 0$ and $h(x) = e^x$. Since $e^x$ is continuous at each $x = b$ you don't have to worry about what particular $b$ results from $lim_{x\rightarrow a} f(x)$ , just as long as the limit exists.

The above result is a theorem. (When people get in the habit of applying a theorem, they begin to think of it as a rule or formula). It requires some effort to prove it. Most calculus textbooks include this theorem. Is it in your course materials?

8. Dec 24, 2014

### Fredrik

Staff Emeritus
\lim_{x\to a}f(x) produces $\lim_{x\to a}f(x)$ or $$\lim_{x\to a}f(x)$$ If you want the former to look like the latter, use \displaystyle like this: {\displaystyle\lim_{x\to a}f(x)}

9. Dec 24, 2014

### Staff: Mentor

Thanks Stephen and Fredrik.

10. Dec 24, 2014

### andyrk

No, I don't think it is. What is this theorem called exactly? Can you give me some URL link where I can study it more deeply?

How would we even get to know what b is until we have evaluated the limit? And for evaluating the limit we need to apply the theorem. But for the theorem we need b. Isn't it ambiguous?

Last edited: Dec 24, 2014
11. Dec 24, 2014

### Staff: Mentor

12. Dec 24, 2014

### andyrk

It has discussion on constant factors inside the limit, while multiplication inside the limit is occurring. It doesn't mention anything about what happens when a constant is present in the exponential form.

13. Dec 24, 2014

### FactChecker

It's not so much a theorem as it is the definition of "continuous". limx->aex = elimx->ax = ea because ex is continuous at a.

Last edited: Dec 24, 2014
14. Dec 24, 2014

### Staff: Mentor

15. Dec 24, 2014

### Stephen Tashi

You can find the theorem by searching on "limit of composite functions" , "composite limit theorem". A link to a video (that I haven't watched myself yet) is http://www.larsoncalculus.com/calc1...he-limit-of-a-composite-function/#content-top

The purpose of a theorem is establish a valid mathematical conclusion from certain "givens". The techniques of working problems, often don't proceed deductively. They often assume the desired conclusion exists and work backwards. The fact that a technique of working problems uses methods that aren't not valid deductively doesn't mean there is any ambiguity about mathematical theorems. It means that the problem solving technique isn't rigorous mathematical deduction.

To evaluate the limit $\lim_{x \rightarrow a} h(f(x))$ , most people would evaluate $lim_{x\rightarrow a} f(x) = b$ and then ( if they were being careful) they would ask themselves if $h(x)$ was continuous at $x = b$.

Last edited: Dec 24, 2014
16. Dec 24, 2014

### DarthMatter

Sorry, of course the last equation should be ${\displaystyle \lim_{x\rightarrow x_0} e^{g(x)}=e^a}$. Thanks for the displaystyle, Fredrik! Also the $B_{\epsilon'}$ should maybe be called by another name. I hope the idea still comes through: You can make the difference between $e^a$ and $e^{g(x)}$ arbitrarily small by making $|g(x)-a|$ 'small enough'.

This I do not understand.

17. Dec 28, 2014

### andyrk

But why would anyone do that? Who tells them to do that? The textbook gives ways to evaluate explicit limits, not implicit. Then how can one simply evaluate limit on $f(x)$ without any valid reason/grounds to do so?

And I think that you are saying to check whether $h(x)$ is continous at $x=a$ because $h(x)$ is within a limit. Had this not been true and had $h(x)$ been outside the limit then? I mean had it been like - $h(\lim_{x \rightarrow a} f(x) )$ instead of $\lim_{x \rightarrow a} h(f(x))$ then would you say that we need not check for whether $h(x)$ is continous at $x=b$?

I am saying this because according to the theorem you have posted, this implies that $\lim_{x \rightarrow a} h(f(x))$ = $h(\lim_{x \rightarrow a} f(x) )$.

And since $h(\lim_{x \rightarrow a} f(x) )$ doesn't require to check whether $h(x)$ is continous at $x=b$ or not, one need not check it. But this differs to what you said.

So had it even been $\lim_{x \rightarrow a} h(f(x))$ it would have been converted to $h(\lim_{x \rightarrow a} f(x) )$ and hence we can argue that even for $\lim_{x \rightarrow a} h(f(x))$ we don't need to check whether $h(x)$ is continous at $x=b$ or not.

And sorry to say, but I didn't understand the proof equally well either.

Last edited: Dec 28, 2014
18. Dec 28, 2014

### Stephen Tashi

I can't understand what you are saying or asking. I suggest you consider some examples.

For example, define the function $h(x)$ as follows:
if $x < 0$ then $h(x) = 0$
if $x \ge 0$ then $h(x) = 2 + x$

Define the function $g(x)$ as $g(x) = x - 5$

Consider $\lim_{x \rightarrow 5} h(g(x))$. This limit does not exist.

Consider $\lim_{x \rightarrow 0} h(g(x))$ This limit exists and is equal to zero.

19. Dec 28, 2014

### andyrk

Yes, you are correct on this one. But what about - $h(\lim_{x \rightarrow 5} g(x))$? Wouldn't this be equal to 2?

20. Dec 28, 2014

### Stephen Tashi

By the composite limit theorem, the limit would be $h(-5) = 0$.