Find the value of the following limit.
![\[ \lim_{x \to 0^+} \frac{x - \sin x}{(x \sin x)^{\frac{3}{2}}}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-6cccf6d7c79d172dc3e884bb16980bbe_l3.png "Rendered by QuickLaTeX.com")
First, since 
we can multiply by this inside the limit without changing the value,
![\begin{align*} \lim_{x \to 0^+} \frac{x - \sin x}{(x \sin x)^{\frac{3}{2}}} &= \lim_{x \to 0^+} \left( \frac{x - \sin x}{(x \sin x)^{\frac{3}{2}}} \cdot \left( \frac{\sin x}{x} \right)^{\frac{3}{2}} \right) \\[9pt] &= \lim_{x \to 0^+} \frac{x - \sin x}{x^3}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ad5d743a494e5c4ae00c1de7518309c6_l3.png "Rendered by QuickLaTeX.com")
Now we can apply L’Hopital’s rule to compute the limit,
![\begin{align*} \lim_{x \to 0^+} \frac{x - \sin x}{x^3} &= \frac{ 1 - \cos x}{3x^2} \\[9pt] &= \lim_{x \to 0^+} \frac{ \sin x}{6x} \\[9pt] &= \lim_{x \to 0^+} \frac{\cos x}{6} \\[9pt] &= \frac{1}{6}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-1b22dbd4510ae3585a522e1cbf66d007_l3.png "Rendered by QuickLaTeX.com")
Since this limit exists, the application of L’Hopital’s rule was justified.