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?
 A: $$\lim_{x\to a}\left(\dfrac{\sin x}{\sin a}\right)^{\dfrac1{x-a}}$$
$$=e^{\lim_{x\to a}\left(\frac{\ln \frac{\sin x}{\sin a}}{x-a}\right)}$$
Now, $\lim_{x\to a}\left(\frac{\ln \frac{\sin x}{\sin a}}{x-a}\right)=\lim_{x\to a}\left(\frac{\ln \sin x -\ln\sin a}{x-a}\right)$ which is of the form $\frac00$
So, applying L'Hospital Rule, $\lim_{x\to a}\left(\frac{\ln \sin x -\ln\sin a}{x-a}\right)=\lim_{x\to a}\left(\frac{\frac{\cos x}{\sin x}}1\right)=\cot a$

$$\lim_{x\to a}\left(\dfrac{\sin x}{\sin a}\right)^{\dfrac1{x-a}}$$
$$=\lim_{x\to a}\left(\left(1+\dfrac{\sin x-\sin a}{\sin a}\right)^{\dfrac{\sin a}{\sin x-\sin a}}\right)^{\dfrac{\sin x-\sin a}{(x-a)\sin a}}$$
$$=\left(\lim_{y\to0}\left(1+y\right)^{\dfrac1y}\right)^{\lim_{x\to a}\dfrac{\sin x-\sin a}{(x-a)\sin a}} \text{( Putting }\dfrac{\sin x-\sin a}{\sin a}=y,\text{  as }x\to a,y\to 0 \text{)}$$ 
Now , $\lim_{x\to a}\dfrac{\sin x-\sin a}{(x-a)}=\lim_{x\to a}\dfrac{\cos x}1$ (applying L'Hospital Rule)
$\implies \lim_{x\to a}\dfrac{\sin x-\sin a}{(x-a)}=\cos a$
$$\implies \lim_{x\to a}\left(\dfrac{\sin x}{\sin a}\right)^{\dfrac1{x-a}}=e^{\dfrac{\cos a}{\sin a}}=e^{\cot a}$$
A: 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
$$\left(\frac{\sin x}{\sin a}\right)^{\frac{1}{x-a}} = \left(\frac{\sin a + (x - a) \cos a + O((x - a)^2)}{\sin a}\right)^{\frac{1}{x-a}} = \left(1 + (x - a) \cot a + O((x - a)^2)\right)^{\frac{1}{x-a}} = \left[\left(1 + (x - a) \cot a + O((x - a)^2)\right)^{\frac{1}{(x-a) \cot a}}\right]^{\cot a} \quad \stackrel{(x \to a)}{\longrightarrow} \quad e^{\cot a}.$$
