Evaluating Limit - $(1-\cos(x^2))/(x^3\sin(x))$

Question

How would you go about evaluating the following limit as $x$ approaches $0$?

$$\lim_{x\to 0}\frac{1-\cos(x^2)}{x^3\sin(x)}$$

Answer 1

$$\frac{1-\cos(x^2)}{x^3\sin(x)}=\frac{2\sin^2\left(\frac{x^2}{2}\right)}{x^3\sin(x)}=\frac{1}{2}\left(\frac{\sin\left(\frac{x^2}{2}\right)}{\frac{x^2}{2}}\right)^2\frac{x}{\sin(x)}$$

then use $\displaystyle \lim_{x\to 0}\frac{x}{\sin(x)}=1$, you get your result.

Answer 2

If you know that $1-\cos x^2$ ~ $ \frac{x^4}{2}$ and $\sin x$ ~ $x$ as $x\to 0$, then it will be helpful.

So the limit is $$=\lim_{x\to 0}\frac{\frac{x^4}{2}}{x^4}=\frac12.$$

Answer 3

You surely know that $$ \lim_{t\to0}\frac{1-\cos t}{t^2}=\frac{1}{2}, \qquad \lim_{t\to0}\frac{\sin t}{t}=1. $$ The first limit tells you that $$ \lim_{x\to0}\frac{1-\cos(x^2)}{x^4}=\frac{1}{2}, $$ so it's just a matter of rewriting: $$ \lim_{x\to 0}\frac{1-\cos(x^2)}{x^3\sin(x)}= \lim_{x\to 0}\frac{1-\cos(x^2)}{x^4}\frac{x}{\sin(x)} $$