To prove the limit of $\frac{x^2+2\cos x-2}{x\sin^3 x}$ at zero is $1/12$ To Prove $$\lim_{x \to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}=\frac{1}{12}$$
I tried with L'Hospital rule but in vain.
 A: Hint. You may use Taylor expansions, as $x \to 0$,
$$
\cos x =1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6)
$$
$$
\sin x =x+O(x^3)
$$ giving
$$
x^2+2\cos x-2=\frac{x^4}{12}+O(x^6)
$$ $$
x\sin^3 x =x^4+O(x^6)
$$ and

$$
\frac{x^2+2\cos x-2}{x\sin^3 x}=\frac{\frac{x^4}{12}+O(x^6)}{x^4+O(x^6)}=\frac{1}{12}+O(x^2)
$$

A: L'Hopital's rule may be used repeatedly until you find an answer:
$$\lim_{x \to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}=\lim_{x \to 0}\frac{2x-2\sin x}{\sin^2 x(\sin x+3x\cos x)}$$
$$= \lim_{x \to 0}\frac{2-2\cos x}{ 6\sin ^2(x)\cos (x) - 3x(\sin^3(x)-2\sin(x)\cos^2(x))}$$
$$= \lim_{x \to 0}\frac{2\sin (x)}{ 3(-3\sin^3 (x) + 6\cos^2(x)\sin(x) + x(2\cos^3(x)-7\cos(x)\sin^2(x)))}$$
$$= \lim_{x \to 0}\frac{-2\cos(x)}{ -24\cos^3(x)+84\sin^2(x)\cos(x)+x(60\sin(x)\cos^2(x)-21\sin^3(x))}$$
$$= \frac{-2\cos(0)}{ -24\cos^3(0)+84\sin^2(0)\cos(0)+0(60\sin(0)\cos^2(0)-21\sin^3(0))}$$
$$=\frac{1}{12}$$
A: We can use the basic limits 
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\sin x}{x}=1,\ \ \ \ \ \ \ \ \
\lim\limits_{x\rightarrow 0}\dfrac{\cos x-1+\frac{1}{2}x^{2}}{x^{4}}=\dfrac{1%
}{24}
\end{equation*}
as follows
\begin{equation*}
\lim\limits_{x\rightarrow 0}\frac{x^{2}+2\cos x-2}{x\sin ^{3}x}%
=\lim\limits_{x\rightarrow 0}\left( \frac{x}{\sin x}\right)
^{3}\lim\limits_{x\rightarrow 0}\left( \frac{\cos x-1+\frac{1}{2}x^{2}}{x^{4}%
}\right) \left( 2\right) =1^{3}\cdot \frac{1}{24}\cdot 2=\frac{1}{12}.
\end{equation*}
The basic limits can be computed by repeated use of L'Hospital's rule. 
A: from 
$$
\cos x = 1 -  \frac{x^2}2+\frac{x^4}{24}-\dots
$$
we have
$$
x^2+2\cos x - 2 = \frac{2x^4}{24}-\dots
$$
and
$$
\sin x = x-... \\
x\sin^3 x = x^4-...
$$
A: Here is how you can do it via a single application of L'Hospital's Rule. The denominator in the expression can be replaced by $x^4$ via the standard limit $\lim_{x\to 0}(\sin x) /x=1$. And the resulting expression can be rewritten as $$\frac{x^2-4\sin^2(x/2)}{x^4}=\frac{t^2-\sin^2t}{4t^4} =\frac{t-\sin t} {4t^3}\cdot\left(1+\frac{\sin t} {t} \right) $$ using the substitution $x=2t$. The first factor tends to $(1/6)(1/4)=1/24$ via single application of L'Hospital's Rule and the second factor tends to $(1+1)=2$ so that the desired limit is $1/12$.
A little algebraic manipulation combined with standard limits is always a great help when applying the L'Hospital's Rule. 
