What I know

If $\lim\limits_{x \to x_0}f(x) := r$ exists, we can create a new function $\tilde f(x) = \begin{cases} f(x) &\text{if }x\in\mathbb{D}\setminus x_0 \\ r & \text{if }x = x_0 \end{cases}$ which is then the continously extended version of $f$.

What my problem is

I am struggling with $\lim\limits_{x\to 0}\frac{2x^3+x^2+x\sin(x)} {(\exp(x)-1)^2}:=r$.

I tried using L'Hôpital's rule, because I noticed that both denominator and numerator would equal to $0$ if I plug in $0$. This unfortunately didn't help at all, because you can derive those expressions as often as you want, without making your life easier.

I deliberately phrased this question in regards of solving continuity problems like this, because I think that calculating the limit in this subtask of an actual first term exam is too hard. There has to be another way of solving this continuity issue, without having to calculate the limit.

If there's no way around finding $r$, then there has to be an obvious trick that I am unaware of.

Help is greatly appreciated!


The limit is the unique value you can give the function at $0$ to make it continuous at $0$. So computing the limit is unavoidable.

You are probably computing badly. Recall that $$ (e^x-1)^2=e^{2x}-2e^x+1 $$ so you can do \begin{align} \lim_{x\to 0}\frac{2x^3+x^2+x\sin(x)}{e^{2x}-2e^x+1} &=\lim_{x\to 0}\frac{6x^2+2x+\sin x+x\cos x}{2e^{2x}-2e^x}\\[6px] &=\lim_{x\to 0}\frac{12x+2+\cos x+\cos x-x\sin x}{4e^{2x}-2e^x}\\[6px] &=\frac{0+2+1+1-0}{4-2}=2 \end{align}

However, you can shorten the computations if you recall that $$ \lim_{x\to0}\frac{\exp x-1}{x}=1 $$ so if your limit exists it is equal to $$ \lim_{x\to0}\frac{2x^3+x^2+x\sin(x)}{x^2}= \lim_{x\to0}\left(2x+1+\frac{\sin x}{x}\right)=2 $$ Or, with Taylor expansion $$ \sin x=x+o(x),\quad \exp x=1+x+o(x) $$ you have $$ \lim_{x\to 0}\frac{2x^3+x^2+x\sin(x)}{(\exp x-1)^2}= \lim_{x\to 0}\frac{2x^3+x^2+x^2+o(x^2)}{(x+o(x))^2}= \lim_{x\to0}\frac{2x^2+o(x^2)}{x^2+o(x^2)}=2 $$

  • $\begingroup$ Thanks! This made it painfully obvious, that you simply have to apply L'Hôpital's rule multiple times without miscalculating. $\endgroup$ – Andreas Schliebitz Feb 12 '16 at 17:12
  • $\begingroup$ @AndreasSchliebitz I added two shorter ways. And, yes, it's better avoiding errors. ;-) $\endgroup$ – egreg Feb 12 '16 at 17:15
  • $\begingroup$ Really appreciate it, great answer! $\endgroup$ – Andreas Schliebitz Feb 12 '16 at 17:16

You can use Taylor series, so the limit become $\lim_\limits{x \to 0}f(x) \simeq \frac{2x^2}{x^2}=2$


$\sin(x)\simeq x$

$e^x \simeq 1+x$


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