# How to evaluate $\lim_{x\to 0} \frac{x^2\sin {\frac{1}{x}}}{\sin x}$

Find the value of $\lim_{x\to 0} \dfrac{x^2\sin {\dfrac{1}{x}}}{\sin x}$.

Below I am showing my attempt at the question:

$x^2\sin {\dfrac{1}{x}}\to 0$ as $x\to 0$ since $\sin {\dfrac{1}{x}}$ is bounded in a neighbourhood of $0$ and $x^2\to 0$ as $x\to 0$.

Hence , we have a $\dfrac{0}{0}$ form.

By L'Hospital's Rule ,$\lim_{x\to 0} \dfrac{x^2\sin {\dfrac{1}{x}}}{\sin x}$ reduces to $\dfrac{2x\sin {\dfrac{1}{x}}-\cos{\dfrac{1}{x}}}{\cos x}$ whose limit can't be computed at $x=0$ since $\lim _{x\to 0} \cos{\dfrac{1}{x}}$ does not exist.

How can I evaluate this correctly?

Will the answer be limit does not exist?

Do excuse me if I am unable to post a good question as this is my first question on MSE

• Apply L'Hospital's rule onto$$\lim_{x\to0}\frac{x^2}{\sin x}$$or you could simply use$$\lim_{x\to0}x=0,\quad\lim_{x\to0}\frac x{\sin x}=1$$and then squeeze:$$\frac{x^2\sin\frac1x}{\sin x}$$(P.S. I think it's a good question, thanks for including your own efforts, thoughts, etc.) – Simply Beautiful Art Dec 27 '17 at 15:19
• @SimplyBeautifulArt;Thank you very much – Learnmore Dec 27 '17 at 15:24
• Incidentally, OP's example shows a subtlety that I'm sure eluded me when I first learned L'Hopital's rule. If f(x)/g(x) has the form 0/0, and if f'(x)/g'(x) approaches a limit, then we can conclude f(x)/g(x) approaches the same limit. However, if f'(x)/g'(x) does not approach a limit, this does not guarantee that f(x)/g(x) also fails to approach a limit. One sometimes informally paraphrases L'Hopital's rule as "f(x)/g(x) has the same limiting behavior as f'(x)/g'(x)" but one needs to be careful with this informal paraphrase. – idmercer Dec 27 '17 at 16:02
• @idmercer: That's indeed the problem with most teaching of the L'Hopital rule, where the teacher either does not know the precise statement or cannot be bothered to get the conditions right, and then students use it as a method to 'reduce' limits, which of course is totally wrong. – user21820 Dec 28 '17 at 3:54

Just observe that $$\left|x^2\sin {\dfrac{1}{x}}\right|\le x^2$$ giving, as $x \to 0$, $$\left|\dfrac{x^2\sin {\dfrac{1}{x}}}{\sin x}\right|\le \frac{x^2}{\left|\sin x\right|}=\left|\frac{x}{\sin x}\right|\times|x|\to 1\times0=0.$$

Hint:

The required limit can be written as: $$\lim_{x \to 0} \frac{\sin \frac{1}{x}}{\frac{1}{x}} \times \frac{x}{\sin x} = 0 \times 1 =0$$

• Thank you very much for making things clear for me! – Learnmore Dec 27 '17 at 15:26

Note that: $$-\frac{x^2}{\sin(x)} \leq \frac{x^2}{\sin x} \sin(\frac{1}{x}) \leq \frac{x^2}{\sin x}$$ So by the squeezing theorem, the limit is $0$.

Bounding the numerator in the way pointed out already $$|x^2\sin(1/x)|\leq x^2$$ and Taylor expanding the denominator yields $$-\frac{x^2}{x+O(x^3)}<\frac{x^2\sin(1/x)}{\sin x}<\frac{x^2}{x+O(x^3)}$$ and you may conclude that the limit is zero by the squeeze theorem.

\begin{align}L&=\lim_\limits{x\to0}\dfrac{x^2\sin\dfrac1x}{\sin x}\\&=\lim_\limits{x\to0}\dfrac x{\sin x}\cdot\lim_\limits{x\to0}x\sin\left(\dfrac1x\right)\\&=1\cdot\lim_\limits{x\to0}x\sin\left(\dfrac1x\right)\\\hline\text{We know}\\-1&\le\sin\left(\dfrac1x\right)\le1\\-x&\le x\sin\left(\dfrac1x\right)\le x\\\text{Since}\\\lim_\limits{x\to0}(-x)&=0=\lim_\limits{x\to0}x\\\text{Hence}\\&\lim_\limits{x\to0} x\sin\left(\dfrac1x\right)=0\end{align} Using this result in $L$, we have $L=0$