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.
 A: 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.
A: Let $x=1+y$, then
$$\lim \limits _{x \rightarrow 1} \dfrac{\sin (\pi x^{\alpha})}{\sin (\pi x^{\beta})}=\lim \limits _{y \rightarrow 0} \dfrac{\sin (\pi (1+y)^{\alpha})}{\sin (\pi (1+y)^{\beta})}=\lim \limits _{y \rightarrow 0} \dfrac{\sin (\pi+\pi\alpha y))}{\sin (\pi+\pi\beta y)}$$
$$=\lim \limits _{y \rightarrow 0} \dfrac{\sin (\pi\alpha y)}{\sin (\pi\beta y)}=\lim \limits _{y \rightarrow 0} \dfrac{\pi\alpha y}{\pi\beta y}=\frac{\alpha}{\beta}$$
A: $$\lim_{x \to 1} \frac{\sin (\pi x^\alpha)}{\sin(\pi x^\beta)} = \lim_{x \to 1} \frac{\sin (\pi x^\alpha-\pi)}{\sin(\pi x^\beta-\pi)}=\lim_{x \to 1} \frac{\sin(\pi x^\alpha - 1)}{\pi x^\alpha - \pi}\frac{\pi x^\beta - \pi}{\sin(\pi x^\beta-\pi)}\frac{x^\alpha-1}{x^\beta-1}$$
Using the fact that $\sin t/t \to 1$ as $t \to 0$ you get that the expeted limit is
$$ \lim_{x \to 1}\frac{x^\alpha-1}{x^\beta-1}.$$
Now use the fact that $\frac{x^\alpha-1}{x-1} \to \alpha$ as $x \to 1$
A: $$\lim \limits _{x \rightarrow 1} \dfrac{\sin (\pi x^{\alpha})}{\sin (\pi x^{\beta})}=\lim \limits _{x \rightarrow 1} \dfrac{\sin (\pi x^{\alpha}+\pi-\pi)}{\sin (\pi x^{\beta}+\pi-\pi)}=\lim_{x \to 1}\frac{\sin (\pi(x^{\alpha}-1))}{\sin (\pi(x^{\beta}-1))}$$
$$=\lim_{x \to 1}\frac{\sin (\pi(x^{\alpha}-1))}{\pi(x^{\alpha}-1)}\times\frac{1}{\frac{\sin \pi(x^{\beta}-1)}{\pi(x^{\beta}-1)}}\times\frac{\pi(x^{\alpha}-1)}{\pi(x^{\beta}-1)}=\lim_{x \to 1} \frac{x^{\alpha}-1}{x^{\beta}-1}$$
$$=\lim_{x \to 1}\frac{(x-1)(x^{\alpha-1}+...+1)}{(x-1)(x^{\beta-1}+...+1)}=\frac{\alpha}{\beta}$$
