1
$\begingroup$

So I'm trying to evaluate limit written in the title without L'Hospital nor series (cause its not introduced in our course),I tried to use this recognized limit $\lim_{x\to0} \frac{\left(e^x−1\right)}x =1$ so our limit equal $\lim_{x\to0} x\left(\frac{x}{e^x−1}\right)$.

I'm not sure how to prove that $\left(\frac{x}{e^x−1}\right) = 1$ Any facts I can use here or other algebraic manipulation I can use to evaluate limit?

$\endgroup$
  • $\begingroup$ Not $x/e^x -1.$ You want $x/(e^x-1).$ $\endgroup$ – zhw. Dec 10 '17 at 21:58
3
$\begingroup$

Simply observe that:

$$\frac{x^2}{e^x-1}=x\frac{x}{e^x-1}\to0\cdot 1=0$$

NOTE

in this case algebric rule for multiplication holds

NOTE

$$\frac{x}{e^x-1}=\frac{1}{\frac{e^x-1}{x}}\to \frac{1}{1}=1$$

$\endgroup$
  • $\begingroup$ How did you know that limit of $(x/e^x-1)$ is same as its reciprocal cause thats where I got stuck. $\endgroup$ – Dota Dec 9 '17 at 22:10
  • $\begingroup$ I update the aswer explaining that point! $\endgroup$ – gimusi Dec 9 '17 at 22:13
  • $\begingroup$ @Dota is it clear now? $\endgroup$ – gimusi Dec 9 '17 at 22:16
  • $\begingroup$ yes, thank you. $\endgroup$ – Dota Dec 9 '17 at 22:19
  • $\begingroup$ You're Welcome! Bye. $\endgroup$ – gimusi Dec 9 '17 at 22:19
1
$\begingroup$

Let $f(x)=e^x$.

Then $f(0)=1$ thus $$\lim_{x \to 0} \frac{e^x-1}{x}=\lim_{x \to 0}\frac{f(x)-f(0)}{x-0}=f'(0)=1$$

So we have: $$\lim_{x \to 0} \frac{x}{e^x-1}=\frac{1}{f'(0)}=1$$

Continue from here..

$\endgroup$
1
$\begingroup$

Just for fun, if you assume the limit exists, here are a couple of ways to show it must be zero.

  1. If $L=\lim_{x\to0}{x^2\over e^x-1}$, then, letting $u=2x$, which also tends to $0$ as $x$ tends to $0$, we have

$${x^2\over e^x-1}={e^x+1\over4}{4x^2\over(e^x+1)(e^x-1)}={e^x+1\over4}{(2x)^2\over e^{2x}-1}={e^x+1\over4}{u^2\over e^u-1}\to{1+1\over4}L={1\over2}L$$

so $L={1\over2}L$, which implies $L=0$.

  1. For $x\gt0$ we have $e^x\gt1$, so ${x^2\over e^x-1}\gt0$, hence $\lim_{x\to0}{x^2\over e^x-1}\ge0$. But for $x\lt0$ we have $e^x\lt1$, so ${x^2\over e^x-1}\lt0$, hence $\lim_{x\to0}{x^2\over e^x-1}\le0$.

Note, neither of these prove that the limit is $0$, they just say it can't be any other real number.

$\endgroup$
0
$\begingroup$

$e^{x}=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+O(x^4)$, so: $$\frac{x^2}{e^x-1}=\frac{x^2}{1+x+\frac{x^2}{2}+\frac{x^3}{3!}+O(x^4)-1}=\frac{x^2}{x+\frac{x^2}{2}+\frac{x^3}{3!}+O(x^4)}=\frac{x}{1+x+\frac{x}{2}+\frac{x^2}{3!}+O(x^3)}\to\frac{0}{1}=0$$

$\endgroup$
0
$\begingroup$

Let $f:x\to e^x$

$\displaystyle\lim_{x\to 0}\; x\cdot\dfrac{x}{e^x-1}=\lim_{x\to 0}\; x\cdot\dfrac{x-0}{e^x-e^0}=\lim_{x\to 0}\; x\dfrac{1}{f'(0)}=\lim_{x\to 0}\; x\dfrac{1}{e^0}=0$

$\endgroup$
0
$\begingroup$

If you know that $e^x \ge 1+x$, then $e^x-1 \ge x$ so $\dfrac{x^2}{e^x-1} \le \dfrac{x^2}{x} = x \to 0$ as $x \to 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.