h) \( \lim _{x \rightarrow 0^{+}}(1+x)^{\cot x} \)

To evaluate the limit \( \lim _{x \rightarrow 0^{+}}(1+x)^{\cot x} \), we can start by rewriting it in a form that is easier to handle. We know that \( \cot x = \frac{\cos x}{\sin x} \). As \( x \to 0^{+} \), \( \cot x \) approaches \( \infty \) because \( \sin x \to 0 \). Thus, we can express the limit as follows: \[ y = (1+x)^{\cot x} \] Taking the natural logarithm on both sides, we have: \[ \ln y = \cot x \cdot \ln(1+x) \] Now, we focus on finding \( \lim_{x \to 0^{+}} \ln y \): \[ \ln y = \frac{\ln(1+x)}{\tan x} \] Next, we can evaluate \( \lim_{x \to 0^{+}} \frac{\ln(1+x)}{\tan x} \). Both the numerator and denominator approach \( 0 \) as \( x \to 0^{+} \), allowing us to apply L'Hôpital's rule: 1. Differentiate the numerator and denominator: - The derivative of \( \ln(1+x) \) is \( \frac{1}{1+x} \). - The derivative of \( \tan x \) is \( \sec^2 x \). 2. Now applying L'Hôpital's rule: \[ \lim_{x \to 0^{+}} \frac{\ln(1+x)}{\tan x} = \lim_{x \to 0^{+}} \frac{\frac{1}{1+x}}{\sec^2 x} = \lim_{x \to 0^{+}} \frac{\cos^2 x}{1+x} \] At \( x = 0 \), this limit evaluates to: \[ \frac{\cos^2(0)}{1+0} = \frac{1}{1} = 1 \] Thus, we find that: \[ \lim_{x \to 0^{+}} \ln y = 1 \] Consequently, taking the exponential of both sides gives: \[ \lim_{x \to 0^{+}} y = e^1 = e \] Therefore, the final answer is: \[ \lim _{x \rightarrow 0^{+}}(1+x)^{\cot x} = e \]