Evaluate $\lim \limits_{x\to 1}\frac{\sin (\pi x^{\alpha})}{\sin (\pi x^{\beta})}$ Evaluate the following limit: $$\lim \limits_{x\to 1}\dfrac{\sin (\pi x^{\alpha})}{\sin (\pi x^{\beta})}$$ without using L'Hospital rule. I suppose that here we can use $\lim \limits_{x\to 0}\dfrac{\sin x}{x}=1$ but I can't use it.
Can anyone help please?
 A: Hint
Replace $x=1+y$ and use the generalized binomial theorem $$x^a=(1+y)^a=1+a y+\frac{1}{2} a\left(a-1\right) y^2+\cdots$$ So $$\sin(\pi x^a)=\sin(\pi+\pi a y+\cdots)=-\sin(\pi a y+\cdots)$$ Now, use the fact that, for small $z$, $\sin(z)\simeq z$.
I am sure that you can take it from here.
A: Notice, $$\lim_{x\to 1}\frac{\sin(\pi x^{\alpha})}{\sin(\pi x^{\beta})}$$
$$=\lim_{x\to 0}\frac{\sin(\pi (1-x)^{\alpha})}{\sin(\pi (1-x)^{\beta})}$$
using binomial expansion: $(1-x)^{\alpha}=1-\alpha x$ & $(1-x)^{\beta}=1-\beta x$ as $x\to 0$, 
$$=\lim_{x\to 0}\frac{\sin\left(\pi(1-\alpha x)\right)}{\sin\left(\pi(1-\beta x)\right)}$$
$$=\lim_{x\to 0}\frac{\sin\left(\pi-\alpha\pi x\right)}{\sin\left(\pi-\beta\pi  x\right)}$$
$$=\lim_{x\to 0}\frac{\sin\left( \alpha \pi x\right)}{\sin\left( \beta \pi x\right)}$$
$$=\lim_{x\to 0}\frac{\alpha}{\beta}\cdot \frac{\sin\left( \alpha \pi x\right)}{\alpha\pi x}\cdot \frac{\beta \pi x}{\sin(\beta \pi x)}$$
$$=\frac{\alpha}{\beta}\cdot \lim_{x\to 0}\frac{\sin\left( \alpha \pi x\right)}{\alpha\pi x}\cdot \lim_{x\to 0}\frac{\beta \pi x}{\sin(\beta \pi x)}$$
$$=\frac{\alpha}{\beta}(1)\cdot (1)=\color{red}{\frac{\alpha}{\beta}}$$
A: It is necessary to understand that $\beta$ must be non-zero for the limit to exist. So let's assume that $\beta \neq 0$. Further if $\alpha = 0$ then the limit is obviously $0$. Hence let's also consider $\alpha \neq 0$. Then we have
\begin{align}
L &= \lim_{x \to 1}\frac{\sin(\pi x^{\alpha})}{\sin(\pi x^{\beta})}\notag\\
&= \lim_{x \to 1}\frac{\sin(\pi - \pi x^{\alpha})}{\sin(\pi - \pi x^{\beta})}\notag\\
&= \lim_{x \to 1}\frac{\sin(\pi(1 - x^{\alpha}))}{\pi(1 - x^{\alpha})}\cdot\frac{\pi(1 - x^{\alpha})}{\pi(1 - x^{\beta})}\cdot\frac{\pi(1 - x^{\beta})}{\sin(\pi(1 - x^{\beta}))}\notag\\
&= \lim_{x \to 1}1\cdot\frac{1 - x^{\alpha}}{1 - x^{\beta}}\cdot 1\notag\\
&= \lim_{x \to 1}\frac{x^{\alpha} - 1}{x - 1}\bigg/\frac{x^{\beta} - 1}{x - 1}\notag\\
&= \frac{\alpha}{\beta}\notag
\end{align}
The following standard limit have been used $$\lim_{t \to 0}\frac{\sin t}{t} = 1,\,\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$
