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 $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$?
 A: 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$$\lim_{x\to0}\left(\frac{\cos x-1}{x\sin x}+\frac{x-\sin x}{x^2\sin x}\right)=-\frac12+\frac16=-\frac13.$$
A: We have
$$\frac {x \cos x - \sin x} {x^2 \sin x}=\frac{\cos x}{\frac{\sin x}x}\cdot\frac{x-\tan x}{x^3}\to1\cdot \left(-\frac13\right)=-\frac13$$
using

*

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*Are all limits solvable without L'Hôpital Rule or Series Expansion
A: The limit in question is equal to the second derivative of the sinc function evaluated at $0$.  That is,
$$\begin{align}
\lim_{x\to 0}\frac{x\cos(x)-\sin(x)}{x^2\sin(x)}&=\lim_{x\to 0}\frac{x\cos(x)-\sin(x)}{x^3}\frac{x}{\sin(x)}\\\\
&=2\lim_{h\to 0}\frac{\text{sinc}(h)-1}{h^2}\\\\
&=2\lim_{h\to 0}\frac{\sin(h)-h}{h^3}\\\\
\end{align}$$
In This Answer, I showed, without use of calculus, that the sine function satisfies the inequality
$$\sin(h)\ge h-\frac16 h^3\tag1$$
In a parallel development, one can show, without calculus, that $\sin(h)\le h-\frac16h^3+\frac1{120}h^5$.  (Alternatively, integrate $(1)$ twice and use $\cos(0)=1$ and $\sin(0)=0$.)
Hence, applying the squeeze theorem, we find that
$$\lim_{x\to 0}\frac{x\cos(x)-\sin(x)}{x^2\sin(x)}=-\frac13$$
A: While searching more easy way, let me suggest one possible way to solve main difficult part of suggested limit. I change denumerator to $x^3$, for simplicity, as it's equivalent $x^2\sin x$
Suppose we know existence of limit. Then
$$L=\lim_{x\to0}\frac{x-\sin x}{x^3} = \lim_{x\to0}\frac{x-3\sin \frac{x}{3}+4 \sin^3 \frac{x}{3}}{x^3}=\\ =\lim_{x\to0}\left(3\frac{\frac{x}{3} - \sin \frac{x}{3}}{x^3} + \frac{4 \sin^3 \frac{x}{3}}{x^3}\right) =\lim_{x\to0}\left(\frac{\frac{x}{3} - \sin \frac{x}{3}}{9\left(\frac{x}{3}\right)^3} + \frac{4 \sin^3 \frac{x}{3}}{x^3}\right)=\frac{L}{9}+\frac{4}{27}$$
From obtained equation $L=\frac{1}{6}$
A: 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, $L= -\frac13$.
