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

To solve the limit \( \lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{x}} \), we can use the fact that as \( x \) approaches 0, \( \sin x \) approaches 0. Therefore, we can rewrite the expression as: \[ (1+\sin x)^{\frac{1}{x}} \to (1+0)^{\frac{1}{0}} = 1^{\infty} \] This indeterminate form suggests we should use the exponential function and natural logarithm for simplification. We can express the limit in logarithmic form: \[ \lim_{x \to 0} \frac{\ln(1 + \sin x)}{x} \] By applying L'Hôpital's rule, we differentiate the numerator and denominator: 1. The derivative of the numerator \( \ln(1 + \sin x) \) is \( \frac{\cos x}{1 + \sin x} \). 2. The derivative of the denominator \( x \) is simply 1. Thus, our limit becomes: \[ \lim_{x \to 0} \frac{\cos x}{1 + \sin x} \] Substituting \( x = 0 \) gives: \[ \frac{\cos(0)}{1 + \sin(0)} = \frac{1}{1} = 1 \] So, we have: \[ \lim_{x \to 0} \frac{\ln(1+\sin x)}{x} = 1 \] Now exponentiating to find the original limit: \[ \lim_{x \to 0} (1+\sin x)^{\frac{1}{x}} = e^{\lim_{x \to 0} \frac{\ln(1+\sin x)}{x}} = e^1 = e \] Hence, the final result is: \[ \lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{x}} = e \]