Question$\lim _{x \rightarrow 0} \frac{1-e^{-x}}{\sin x}$

Studdy Solution
Evaluate the limit using the derivatives:
$$\lim_{x \to 0} \frac{e^{-x}}{\cos x}$$
Substitute $x = 0$:
$$\frac{e^0}{\cos 0} = \frac{1}{1} = 1$$
The value of the limit is:
$$\boxed{1}$$