Trying to find $\lim_{x \to 0} \frac{x - \sin x}{(x \sin x)^{(3/2)}}$ using L'Hopital's I'm trying to use L'Hopital's rule to calculate:
$$\lim_{x \to 0^+} \dfrac{x - \sin x}{(x \sin x)^{(3/2)}}$$
Taking a couple of derivatives of the denominator gets quite nasty, so I'd like to find a simpler way to do it.
I would like to make a change of variable, say, $t = \sqrt{x \sin x}$, to get a $t^3$ in the denominator.  Unfortunately, that leaves me with problems in the numerator.  Maybe there is some other manipulation or some trig identity that simplifies things that I am missing?  This shouldn't be a difficult problem, but I can't seem to find a slick way to do it.
The answer is given as $\frac{1}{6}$.  Thanks for your help.
 A: It might be appropriate to use the limits or infinitesimal properties of $\sin x$ at $x=0$
$${x- \sin x} \sim \frac{x^3}{3!}$$
Then
$$\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sin x - x}}{{{{\left( {x\sin x} \right)}^{\frac{3}{2}}}}} = \frac{1}{6}\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{x^3}}}{{{{\left( {x\sin x} \right)}^{\frac{3}{2}}}}}$$
and
$$\frac{1}{6}\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{x^3}}}{{{{\left( {x\sin x} \right)}^{\frac{3}{2}}}}} = \frac{1}{6}{\left( {\mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{{\sqrt {x\sin x} }}} \right)^3} = \frac{1}{6}{\left( {\mathop {\lim }\limits_{x \to {0^ + }} \sqrt {\frac{x}{{\sin x}}} } \right)^3}$$
Using the known limit $$\frac{x}{{\sin x}} \to 1$$
one has
$$\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sin x - x}}{{{{\left( {x\sin x} \right)}^{\frac{3}{2}}}}} = \frac{1}{6}$$
A: First answer:
Try squaring the function first. The limit of $f^2$ will exist, and may be less messy via L'Hospital, although it will take more (twice as many) iterations. Then you'll need to decide which square root to take. But surely this $f$ is positive for small $x$.
Better answer: Assuming you know $\lim_{x\to0}\frac{\sin(x)}{x}=1$ (which you certainly do if LH is available), use that to your advantage to remove $\sin$ from the denominator:
$$
\begin{align}
\lim_{x\to0^+}\frac{x-\sin x}{(x\,\sin x)^{3/2}}&=\lim_{x\to0^+}\frac{x-\sin x}{(x\,\sin x)^{3/2}}\cdot\left(\frac{\sin x}{x}\right)^{3/2}\\&=\lim_{x\to0^+}\frac{x-\sin x}{x^3}
\end{align}
$$
Now L'Hospital is a cake walk.
A: Instead of taking multiple derivatives, maybe stop after the taking one derivative.  Then think about what to do with the $1-\cos x$ in the numerator.  Perhaps you know what
$$
\lim_{x\rightarrow 0}\frac{1-\cos x}{x^2}
$$
is?  If not, try multiplying numerator and denominator by $1+\cos x$.
