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
You need to prove that,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.
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.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.
Indeed, the basics you should know. You did understand the latestest discussion about determinants?Originally Posted by Quick
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.
I don't think you can assume $\displaystyle \ln$ is continuous around 1 , as if you do you might asOriginally Posted by mannyfold
well have assumed $\displaystyle \exp$ continuous at 1, then you could just have plugged
in the value to evaluate the limit.
If the function is continuous at the point in question then it is strictlyQuick, 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.