I am preparing for an entrance exam and have come across the following problem:

Prove that the limit as x --> 1 of e^x = e.

Any help would be greatly appreciated.

Thanks

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- Aug 12th 2006, 06:09 PMmannyfoldEpsilon-Delta Limit Proof
I am preparing for an entrance exam and have come across the following problem:

Prove that the limit as x --> 1 of e^x = e.

Any help would be greatly appreciated.

Thanks - Aug 12th 2006, 06:13 PMQuick
(Don't listen to me as I'm asking my own question)

wouldn't it just be possible to substitute 1 for x into the equation? It's quite obvious that e^1 equals e

(would limits be a reasonable question on quicklopedia?) - Aug 12th 2006, 06:53 PMThePerfectHackerQuote:

Originally Posted by**mannyfold**

For any $\displaystyle \epsilon>0$ there is a $\displaystyle \delta>0$ such as,

$\displaystyle |x-1|<\delta$ (1) is in domain and,

$\displaystyle |e^x-e|<\epsilon$ (2)

Note, inequality (2) is,

$\displaystyle e|e^{x-1}-1|<\epsilon$

But from inequality (1),

$\displaystyle 0<|x-1|<\delta$ implies,

$\displaystyle 1=e^0<e^{|x-1|}<e^{\delta}$ implies,

$\displaystyle 0<e^{|x-1|}-1<e^{\delta}-1$ implies,

$\displaystyle |e^{x-1}-1|<|e^{|x-1|}-1|<|e^{\delta}-1|$

So, if I can make inequality,

$\displaystyle e|e^{\delta}-1|<\epsilon$ true then I can definetly make,

$\displaystyle e|e^{x-1}-1|<\epsilon$ true because it is less (transitive property). From here can can show which $\displaystyle \delta$ to chose.

You have,

$\displaystyle e|e^{\delta}-1|<\epsilon$

$\displaystyle |e^{\delta}-1|<\epsilon/e$

$\displaystyle -\epsilon/e<e^{\delta}-1<\epsilon/e$

Work with positive only,

$\displaystyle e^{\delta}-1<\epsilon/e$

$\displaystyle e^{\delta}<1+\epsilon/e$

$\displaystyle \delta<\ln (1+\epsilon/e)$ ---> Leads to problen.

Thus, chosing this delta will give the desired result.

Okay, I have demonstrated with delta epsilon but I said it leads to a propblem. The entire concept of exponential functions and their properties are based on countinuity. Thus, then I cannot prove that they are countinous using the fact that they are countinous. Thus, I do not see how some one can ask you to prove such as problem.

---

Quote:

Originally Posted by**Quick**

Whenver you can do that to a function that is called,**countinous**and much of math it devoted to this beautiful concept. This is a demonstration of how a scientist thinks and how a mathematician thinks. A scientist understands the limit as what happens to a function as it approaches a number. A mathematician, though the same idea, looks completely different. That is what delta-epsilon is doing. But before you get into that you need to have the concept of the limit down.

Quote:

Originally Posted by**Quick**

- Aug 13th 2006, 05:19 AMmannyfold
Many thanks, PerfectHacker, that is a big help. Generally, I am pretty good at proving such things for polynomials, but the exponential case is a bit more confusing. And even some types of non-polynomials require the mean-value theorem, but it just didn't apply here.

Late last night, I came up with a solution similar to yours (involving ln ). I see what you say about the problem, but since you are using a*different*function, albeit the inverse of e^x, I don't think it is a problem. You need only assume ln is continuous around 1. Also, a possible problem would be if the argument of the ln was < 1 which is not the case here.

Quick, thanks, but substitution is the standard way taught in Calculus, but is not strictly correct. To prove a limit means to employ the epsilon-delta method. - Aug 13th 2006, 05:52 AMCaptainBlackQuote:

Originally Posted by**mannyfold**

well have assumed $\displaystyle \exp$ continuous at 1, then you could just have plugged

in the value to evaluate the limit.

Quote:

Quick, thanks, but substitution is the standard way taught in Calculus, but is not strictly correct. To prove a limit means to employ the epsilon-delta method.

correct, its one of the things implied by continuity.