(c) \( \lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{x}} \)

To evaluate the limit \( \lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{x}} \), we can use the fact that as \( x \to 0 \), \( \sin x \) behaves like \( x \). Thus, we can rewrite \( 1 + \sin x \) as \( 1 + x \) for small values of \( x \). We can then express the limit in exponential form: \[ \lim_{x \to 0} (1 + \sin x)^{\frac{1}{x}} = \lim_{x \to 0} e^{\frac{1}{x} \ln(1 + \sin x)} \] Using the approximation \( \ln(1 + u) \approx u \) when \( u \) is close to 0, we find: \[ \ln(1 + \sin x) \approx \sin x \approx x \quad \text{as } x \to 0 \] Therefore, we can rewrite our expression: \[ \frac{1}{x}\ln(1 + \sin x) \approx \frac{1}{x} \cdot x = 1 \] Now substituting this back into our limit gives: \[ \lim_{x \to 0} e^{\frac{1}{x} \ln(1 + \sin x)} = e^{1} = e \] Thus, the limit evaluates to: \[ \lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{x}} = e \]