Limit calculation, calculus I am trying to find 

$$  \lim_{x\to 0} \frac{e^{\alpha x} -e^{\beta x}}{\sin(\alpha x) + \sin(\beta x) }$$

but dont have a clue where to start, could someone give me a hint please ?
Since $\alpha \neq -\beta$

$$e^{\alpha x} = 1 + \alpha x + o(x)$$
$$e^{\beta x} = 1 + \beta x + o(x)$$
$$\sin(\alpha x) = \alpha x + o(x)$$
$$\sin(\beta x) = \beta x + o(x)$$
So
$$ \frac{e^{\alpha x} -e^{\beta x}}{\sin(\alpha x) + \sin(\beta x) } = \dfrac{1 + \alpha x - 1- \beta x +o(x)}{\alpha x + \beta x +o(x) } = \dfrac{\alpha - \beta + \dfrac{o(x)}{x}}{\alpha + \beta + \dfrac{o(x)}{x}} \to \dfrac{\alpha - \beta}{\alpha + \beta}$$

Alternatively, use 

$$\frac{e^{\alpha x} -e^{\beta x}}{\sin(\alpha x) + \sin(\beta x) } 
=\frac{\dfrac{e^{\alpha x} - 1}{\alpha x}\alpha-\dfrac{e^{\beta x} - 1}{\beta x}\beta}{\dfrac{\sin(\alpha x)}{\alpha x }\alpha + \dfrac{\sin(\beta x)}{\beta x}\beta }$$

and $\lim \dfrac{e^{\alpha x} - 1}{\alpha x} = \lim \dfrac{e^{\beta x} - 1}{\beta x} = \lim \dfrac{\sin(\alpha x)}{\alpha x } = \lim \dfrac{\sin(\beta x)}{\beta x} = 1 $
Using Wolfram Alpha:

lim (e^(alpha x)-e^(beta x))/(sin(alpha x)+sin(\beta x)) as x->0

Answer:

(alpha-beta)/(alpha+beta)

http://www.wolframalpha.com/input/?i=lim+%28e%5E%28alpha+x%29-e%5E%28beta+x%29%29%2F%28sin%28alpha+x%29%2Bsin%28%5Cbeta+x%29%29+as+x-%3E0