My textbook evaluates the following limit as being equal to $0$ by dividing all the terms by $e^x$.

$$\lim_{x\to 0} \frac{x^2}{e^x-1}$$

However, I would like to argue that this limit does not exist, consider

$$\lim_{x\to 0^+} \frac{0+}{1^+-1} = \lim_{x\to 0^+} \frac{0+}{0+} = \infty $$

Now consider $$\lim_{x\to 0^-} \frac{0+}{1^--1} = \lim_{x\to 0^+} \frac{0+}{0-} = -\infty $$

The left side and right side limits are not equal, therefore the limit cannot exist. Is my conclusion correct?

  • 2
    $\begingroup$ Your conclusion is not correct. You cannot simply substitute into the expression what the variable approaches (unless certain conditions are met). In this case, you end up with $0/0$ which is undefined. This means you might end up with a finite limit, or it might be an infinite limit. $\endgroup$ – Clayton Nov 8 '18 at 19:35
  • 2
    $\begingroup$ I think your approach as well the book's approach are wrong. Dividing each term by $e^x$ does not help at all. The right approach is use the well known limit $\lim_{x\to 0}\dfrac {e^x-1}{x}=1$. $\endgroup$ – Paramanand Singh Nov 8 '18 at 21:02

Your argument is wrong. To convince you about it, try to compute $$\lim_{x\to 0} \frac{x^2}{x}$$ With your argument, we would deduce that the above limit doesn't exist.

However, I think we agree that $$\lim_{x\to 0} \frac{x^2}{x}=\lim_{x\to 0} x\frac{x}{x}=\lim_{x\to 0} x=0.$$


As mentioned before $\frac{0}{0}$ is indeterminate, but if you don't want evaluate using the technique mentioned by you, you can also use the power series expansion of $e^x$ to evaluate the limit

$$\lim_{x \rightarrow 0}\frac{x^2}{e^x - 1}=\lim_{x \rightarrow 0}\frac{x^2}{\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\right) - 1}=\lim_{x \rightarrow 0}\frac{x^2}{x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots}=\lim_{x \rightarrow 0}\frac{x^2}{x\left(1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots \right)}=\lim_{x \rightarrow 0}\frac{x}{1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots}=\frac{0}{1}=0$$

  • $\begingroup$ does this really answer the question of OP? $\endgroup$ – Surb Nov 8 '18 at 20:55
  • 3
    $\begingroup$ No, I wanted to give an alternate solution. But yours does $\endgroup$ – Naweed G. Seldon Nov 8 '18 at 20:56
  • 2
    $\begingroup$ ok :), you get my +1 for this totally honest comment :) $\endgroup$ – Surb Nov 8 '18 at 20:58

As other answers and comments mentioned, the answer is NO. Look at the table/graph:

$$\begin{array}{r|r|r|r|r|r} x&x^2&e^x-1&&x&x^2&e^x-1\\ \hline -1&1&-0.6&&1&1&1.718282\\ -0.1&0.01&-0.09516&&0.1&0.01&0.105171\\ -0.01&0.0001&-0.00995&&0.01&0.0001&0.010050\\ -0.001&0.000001&-0.00100&&0.001&0.000001&0.001001\\ -0.0001&0.00000001&-0.00010&&0.0001&0.00000001&0.000100\\ \end{array}$$ Note that the function $y=x^2$ approaches $0$ faster than the function $y=e^x-1$, hence their ratio will also approach $0$.

$\hspace{5cm}$enter image description here

Note that the $y\to 0^+$ as $x\to 0^+$ and $y\to 0^-$ as $x\to 0^-$.

Also note the estimate: $$e^x>x+1 \Rightarrow e^x-1>x \Rightarrow (e^x-1)^2>x^2, -1<x<1;\\ \color{red}{e^x-1}=\frac{(e^x-1)^2}{e^x-1}<\color{red}{\frac{x^2}{e^x-1}}<\color{red}0, -1<x<0;\\ \color{red}0<\color{red}{\frac{x^2}{e^x-1}}<\frac{(e^x-1)^2}{e^x-1}=\color{red}{e^x-1}, 0<x<1.$$ Now take limit: $$\lim_{x\to 0^-} e^x-1\le \lim_{x\to 0^-} \frac{x^2}{e^x-1}\le \lim_{x\to 0^-} 0 \Rightarrow \lim_{x\to 0^-} \frac{x^2}{e^x-1}=0;\\ \lim_{x\to 0^+} 0\le \lim_{x\to 0^+} \frac{x^2}{e^x-1}\le \lim_{x\to 0^+} e^x-1 \Rightarrow \lim_{x\to 0^+} \frac{x^2}{e^x-1}=0.$$


No your way and conclusion is wrong, the limit is in an indeterminate form $\frac 0 0$ and we can't conclude nothing from here.

Also the hint given in your book seems not conlcusive since dividing by $e^x$ we have


which is again in the same indeterminate form $\frac 0 0$ and then we can't conclude nothing from here.

To solve it, the key point is refer to the standard limit

$$\lim_{x\to 0} \frac{e^x-1}{x}=1$$

and from here we can proceed dividing by $x$ (that's maybe justify a typo in your book) or as an alternative simply noting that

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


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.