<s>[INST] Find limits of $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor Expansion. 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 $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$?
 [/INST] Denote $L$ the existing limit. Then, express it as
\begin{align}
L=\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}
&=  \lim _{x \to 0}\frac {x (2\cos^2\frac x2 -1) -2 \sin \frac x2 \cos\frac x2} {2x^2 \sin \frac x2 \cos\frac x2}\\
 &= \lim _{x \to 0} \frac {x (\cos^2\frac x2-1) +2 \cos\frac x2(\frac x2\cos \frac x2- \sin\frac x2)} {2x^2 \sin \frac x2 \cos\frac x2}\\
&= - \lim _{x \to 0}\frac{\sin\frac x2}{\frac x2} \frac1{4\cos\frac x2}
+ \lim _{x \to 0} \frac{\frac x2\cos \frac x2- \sin\frac x2} {4(\frac x2)^2 \sin \frac x2} \\
&= -\frac14+\frac14L
\end{align}
Thus, $L= -\frac13$.
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