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?
Best Answer
Hint:
$$1-\cos x = 2\sin^2 \frac{x}{2}$$
also:
$$\frac{2\sin \frac{x}{2}}{x - 2\sin \frac{x}{2}\cos \frac{x}{2}} $$
$$\implies \frac{2}{\frac{1}{2} \frac{\frac{x}{2}}{\sin \frac{x}{2}} -{2}\cos \frac{x}{2}} $$
where:
$$\lim_{\alpha \rightarrow 0} \frac{\sin \alpha}{\alpha} = 1$$
here consider: $\alpha = \frac{x}{2}.$
Then, it's ready for replacing $x \rightarrow 0$!