<s>[INST] Evaluating $\lim_{x \rightarrow 0} \frac{\sin(\pi x)(1-\cos(\pi x)}{x^2\sin(x)}$ without L'hôpital's rule I need help finding this limit:
$$\lim_{x \rightarrow 0}  \frac{\sin(\pi x)(1-\cos(\pi x))}{x^2\sin(x)}$$
I've used L'Hôpital's rule and the solution is $\pi^3/2$. However I'm asked to solved it without using it and I'm stuck since all the trig tricks I can think of are not making the limit any simpler. So what can I do to solve this?
 [/INST] We use the result that $$\lim_{x \to 0}\frac{\sin x}{x}=1$$
from this result we also have, $$\lim_{x \to 0}\frac{\sin \pi x}{\pi x}=1\text{ and } \lim_{x \to 0} \frac{\sin (\frac{\pi x}{2})^2}{(\frac{\pi x}{2})^2}=1$$
Now,$$\lim_{x \to 0}\frac{\sin \pi x \cdot(1-\cos\pi x)}{x^2 \cdot \sin x}=\lim_{x \to 0}\frac{\sin \pi x \cdot(2\sin (\frac{\pi x}{2})^2)}{x^2 \cdot \sin x}\\=\lim_{x \to 0} \frac{\sin \pi x}{\pi x}\cdot\pi\cdot\frac{x}{\sin x}\cdot\frac{\pi ^2}{4}\cdot2\cdot\frac{\sin (\frac{\pi x}{2})^2)}{(\frac{\pi x}{2})^2}=\frac{\pi ^3}{2}\lim_{x \to 0} \frac{\sin \pi x}{\pi x}\cdot\frac{x}{\sin x}\cdot\frac{\sin (\frac{\pi x}{2})^2}{(\frac{\pi x}{2})^2}=\frac{\pi ^3}{2}$$
</s>