# how to evaluate $\lim_{x\to0} (x^2)/(e^x-1)$ without L'Hospital

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?

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

## 6 Answers

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

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

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

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..

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.

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

Let $f:x\to e^x$

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

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 $x \to 0$.