limit of trigonometric function How to find the limit of this question 
$$\lim_{x \rightarrow a} \left( \frac{\sin x}{\sin a} \right)^{\frac{1}{x-a}}$$ where $a \neq k\pi$ with k an integer.
We can write this as : $$\exp\left({\lim_{x\rightarrow a}\dfrac{1}{x-a}\left(\dfrac{\sin x}{\sin a} \right)} \right)$$
How to proceed further?

If it is allowed, you can also use Taylor series around $a$:
$$\sin x = \sin a + (x - a) \sin' a + O((x - a)^2) = \sin a + (x - a) \cos a + O((x - a)^2).$$
Plugging this in, we get
$$\left(\frac{\sin x}{\sin a}\right)^{\frac{1}{x-a}} = \left(\frac{\sin a + (x - a) \cos a + O((x - a)^2)}{\sin a}\right)^{\frac{1}{x-a}} = \left(1 + (x - a) \cot a + O((x - a)^2)\right)^{\frac{1}{x-a}} = \left[\left(1 + (x - a) \cot a + O((x - a)^2)\right)^{\frac{1}{(x-a) \cot a}}\right]^{\cot a} \quad \stackrel{(x \to a)}{\longrightarrow} \quad e^{\cot a}.$$
