# Applying L'hopital rule if limit does not exist in the numerator?

I want to find

$$\lim \limits_{x \to 0} \frac{x^2\sin\left(\frac{1}{x}\right)}{\sin(x)}.$$

I know that the numerator and denominator equal to zero if I plug in $x=0$. Can I use L'Hopital's rule even though the limit of $\sin\frac{1}{x}$ does not exist? Because to use L'Hopital's rule both the denominator and the numerator have to be differentiable?

• – lab bhattacharjee Mar 1 '17 at 13:08
• But the limit does exist in the numerator; $x^2$ tends to zero, and $\sin\frac1x$ is bounded, so the product tends to zero. Remember, the limit of a product can exist even if the individual limits of the factors don't exist. – MPW Mar 1 '17 at 13:24

## 2 Answers

HINT, since $\sin\left(\frac{1}{x}\right)$ is bounded and $\lim_{x\to0}\frac{x^2}{\sin\left(x\right)}=0$, we can conclude......!

You can use hopital because both the numerator and denunerator are differentiable and they both tend to $0$ (to prove that the numerator does tend to $0$, though, you can't just plug $x=0$ in because $\sin \frac 1x$ is not defined in $0$

Another way would be to use Taylor expansions and write $$\sin x \sim x$$ so that your expression is asymptotically equivalent to $x \sin \frac 1x$, which tend to $0$; hence also your original limit is equal to $0$