TITLE: L'Hospital's rule help
QUESTION [5 upvotes]: I tried using L'Hospital's rule to find the limit as $x$ tends to $0$ for the following function:
$$f(x) = \frac{(1 - \cos x)^{1.5}}{x - \sin x}$$
I tried differentiating the top and the bottom and I can do it many times but it still gives the denominator with zero meaning I have to differentiate again and again (mainly because a  $(1 - \cos x)^{0.5}$  always manages to find its way into the numerator.)
Is there a better way of solving this limit question?

REPLY [4 votes]: It appears that you mean $(1 - \cos x)^{3/2}$ in the numerator. Please reflect this by editing your question.I have edited the question to reflect this.
In this case it can be shown that the limit does not exist as left and right hand limits are different. Doing this requires the use of standard limit $$\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2}\tag{2}$$ and the use of L'Hospital's Rule once.
Thus we have for $x \to 0^{+}$
\begin{align}
L &= \lim_{x \to 0^{+}}\frac{(1 - \cos x)^{3/2}}{x - \sin x}\notag\\
&= \lim_{x \to 0^{+}}\left(\frac{1 - \cos x}{x^{2}}\right)^{3/2}\cdot\frac{x^{3}}{x - \sin x}\notag\\
&= \frac{1}{2\sqrt{2}}\lim_{x \to 0^{+}}\frac{x^{3}}{x - \sin x}\notag\\
&= \frac{1}{2\sqrt{2}}\lim_{x \to 0^{+}}\frac{3x^{2}}{1 - \cos x}\text{ (via L'Hospital's Rule)}\notag\\
&= \frac{3}{\sqrt{2}}\notag
\end{align}
The limit as $x \to 0^{-}$ is $-3/\sqrt{2}$ and this can be shown by putting $x = -t$ and then using the above limit evaluation.