Compute limit of a function without using L' Hôpital

Question

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.

Answer 1

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.

Answer 2

$$\lim \limits _{x \rightarrow 1} \dfrac{\sin (\pi x^{\alpha})}{\sin (\pi x^{\beta})}=\lim \limits _{x \rightarrow 1} \dfrac{\sin (\pi x^{\alpha}+\pi-\pi)}{\sin (\pi x^{\beta}+\pi-\pi)}=\lim_{x \to 1}\frac{\sin (\pi(x^{\alpha}-1))}{\sin (\pi(x^{\beta}-1))}$$ $$=\lim_{x \to 1}\frac{\sin (\pi(x^{\alpha}-1))}{\pi(x^{\alpha}-1)}\times\frac{1}{\frac{\sin \pi(x^{\beta}-1)}{\pi(x^{\beta}-1)}}\times\frac{\pi(x^{\alpha}-1)}{\pi(x^{\beta}-1)}=\lim_{x \to 1} \frac{x^{\alpha}-1}{x^{\beta}-1}$$ $$=\lim_{x \to 1}\frac{(x-1)(x^{\alpha-1}+...+1)}{(x-1)(x^{\beta-1}+...+1)}=\frac{\alpha}{\beta}$$

Answer 3

Let $x=1+y$, then $$\lim \limits _{x \rightarrow 1} \dfrac{\sin (\pi x^{\alpha})}{\sin (\pi x^{\beta})}=\lim \limits _{y \rightarrow 0} \dfrac{\sin (\pi (1+y)^{\alpha})}{\sin (\pi (1+y)^{\beta})}=\lim \limits _{y \rightarrow 0} \dfrac{\sin (\pi+\pi\alpha y))}{\sin (\pi+\pi\beta y)}$$

$$=\lim \limits _{y \rightarrow 0} \dfrac{\sin (\pi\alpha y)}{\sin (\pi\beta y)}=\lim \limits _{y \rightarrow 0} \dfrac{\pi\alpha y}{\pi\beta y}=\frac{\alpha}{\beta}$$

Answer 4

$$\lim_{x \to 1} \frac{\sin (\pi x^\alpha)}{\sin(\pi x^\beta)} = \lim_{x \to 1} \frac{\sin (\pi x^\alpha-\pi)}{\sin(\pi x^\beta-\pi)}=\lim_{x \to 1} \frac{\sin(\pi x^\alpha - 1)}{\pi x^\alpha - \pi}\frac{\pi x^\beta - \pi}{\sin(\pi x^\beta-\pi)}\frac{x^\alpha-1}{x^\beta-1}$$

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 $x \to 1$