Find the limit $\displaystyle \lim _{x \to 0} \frac {x \cos x – \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor expansion.
My Try
$\displaystyle =\lim _{x \to 0} \frac {\cos x – \frac{\sin x}{x}} {x \sin x}$
$=\frac {\displaystyle\lim _{x \to 0}\cos x – \lim _{x \to 0}\frac{\sin x}{x}} {\displaystyle\lim _{x \to 0}x \sin x}$
$=\frac{1-1}{0}$
But still I end up with $\frac00$
Any hint for me to proceed would be highly appreciated.
P.S: I did some background check on this question on mathstack and found they have solved this with l'Hopital's rule and the answer seems to be $\frac{-1}{3}$.
What is $\lim _{x \to 0} \frac {x \cos x – \sin x} {x^2 \sin x}$?
Best Answer
Famously $\lim_{x\to0}\frac{\sin x}{x}=1$ can be proved without such techniques, and implies $\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12$. With a little more effort (e.g. by approximating a circular arc as a parabola), you can also show $\lim_{x\to0}\frac{x-\sin x}{x^3}=\tfrac16$. So $\lim_{x\to0}\frac{x-\sin x}{x^2\sin x}=\tfrac16$ and$$\lim_{x\to0}\left(\frac{\cos x-1}{x\sin x}+\frac{x-\sin x}{x^2\sin x}\right)=-\frac12+\frac16=-\frac13.$$