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
    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. – Clayton Nov 8 at 19:35
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    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 $\lim_{x\to 0}\dfrac {e^x-1}{x}=1$. – Paramanand Singh Nov 8 at 21:02
up vote 5 down vote accepted

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

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

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

$$\frac{x^2}{e^x-1}=\frac{\frac{x^2}{e^x}}{1-\frac1{e^x}}$$

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

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

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