Compute limit of a function without using L' Hôpital Compute $\lim \limits _{x \rightarrow 1} \dfrac{\sin (\pi x^{\alpha})}{\sin (\pi x^{\beta})}$ without using L' Hôpital's rule.
It is obvius that the value is $\dfrac{\alpha}{\beta}$ by using L' Hôpital.

If $\alpha\ne0$, the function $x\mapsto \pi x^\alpha-\pi$ is invertible in a neighborhood of $1$, with continuous inverse; of course $\lim_{x\to 1}(\pi x^\alpha-\pi)=0$. Then, setting $t=\pi x^\alpha-\pi$,
$$
\lim_{x\to 1}\frac{\sin(\pi x^\alpha)}{\pi x^\alpha-\pi}=
\lim_{t\to 0}\frac{\sin(t+\pi)}{t}=-1.
$$
So you can rewrite your limit as
$$
\lim_{x\to 1}
\frac{\sin(\pi x^\alpha)}{\pi x^\alpha-\pi}
\frac{\pi x^\beta-\pi}{\sin(\pi x^\beta)}
\frac{\pi x^\alpha-\pi}{\pi x^\beta-\pi}=
\lim_{x\to 1}\frac{\pi x^\alpha-\pi}{\pi x^\beta-\pi}=
\lim_{x\to 1}\frac{x^\alpha-1}{x^\beta-1}
$$
provided this last limit exists (which it does). Now it should be easy, by rewriting the limit as
$$
\lim_{x\to 1}\frac{x^\alpha-1}{x-1}\frac{x-1}{x^\beta-1}.
$$
The cases when $\alpha=0$ or $\beta=0$ should be easy to analyze.
