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?
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,$$\lim_{x \to 0}\frac{\sin \pi x \cdot(1-\cos\pi x)}{x^2 \cdot \sin x}=\lim_{x \to 0}\frac{\sin \pi x \cdot(2\sin (\frac{\pi x}{2})^2)}{x^2 \cdot \sin x}\\=\lim_{x \to 0} \frac{\sin \pi x}{\pi x}\cdot\pi\cdot\frac{x}{\sin x}\cdot\frac{\pi ^2}{4}\cdot2\cdot\frac{\sin (\frac{\pi x}{2})^2)}{(\frac{\pi x}{2})^2}=\frac{\pi ^3}{2}\lim_{x \to 0} \frac{\sin \pi x}{\pi x}\cdot\frac{x}{\sin x}\cdot\frac{\sin (\frac{\pi x}{2})^2}{(\frac{\pi x}{2})^2}=\frac{\pi ^3}{2}$$
The limit can not be evaluated without the knowing that $\frac{\sin x}{x}\to 1$ as $x\to 0$. If you do not know that, then give up. I shall suppose that you know that limit. Note that $1-\cos x=2\sin^2 \frac{x}{2}$, we know that $$\lim_{x \rightarrow 0} \frac{\sin(\pi x)(1-\cos(\pi x)}{x^2\sin(x)}=\lim_{x\to 0}\frac{\sin\pi x}{x}\frac{x}{\sin x}\frac{1-\cos\pi x}{x^2}=\lim_{x\to 0}\frac{\sin\pi x}{x}\frac{x}{\sin x}\frac{2\sin^2 \frac{x}{2}}{x^2}$$ and you get your answer.
