Find limit of function

Question

I need find limit without L'Hospital rule. $\lim _{x\to 0}\left(\frac{1-e^{-x}}{sinx}\right)$ I dont know how transform expression.

Answer 1

With Taylor $$\lim _{x\to 0}\left(\dfrac{1-e^{-x}}{\sin x}\right) = \lim _{x\to 0} \dfrac{1-\frac{1}{1+x}}{x}=\lim _{x\to 0} \dfrac{\frac{x}{1+x}}{x}=\lim _{x\to 0} \dfrac{1}{1+x}=1$$

or

$\dfrac{1-e^{-x}}{\sin x} = \dfrac{\dfrac{e^{-x} -1}{-x}}{\dfrac{\sin x}{x}}\underset{x\to 0}{\longrightarrow} 1$

Answer 2

Hint:

1) $\lim_{x \rightarrow 0} \dfrac{e^x -1}{x}=1.$

2)$\lim_{ x \rightarrow 0}\dfrac{\sin x}{x} =1$.

Can you take it from here?