1. compute lim(x→0)(1 + 2x)^(cot x).

Explanation:

Step1: Let (y=(1 + 2x)^{\cot x})

Take the natural - logarithm of both sides: (\ln y=\cot x\ln(1 + 2x)=\frac{\ln(1 + 2x)}{\tan x})

Step2: Use L'Hopital's rule

As (x\rightarrow0), (\frac{\ln(1 + 2x)}{\tan x}) is in the (\frac{0}{0}) form. Differentiate the numerator and denominator. The derivative of (\ln(1 + 2x)) with respect to (x) is (\frac{2}{1 + 2x}), and the derivative of (\tan x) with respect to (x) is (\sec^{2}x). So (\lim_{x\rightarrow0}\frac{\ln(1 + 2x)}{\tan x}=\lim_{x\rightarrow0}\frac{\frac{2}{1 + 2x}}{\sec^{2}x})

Step3: Evaluate the limit

Substitute (x = 0) into (\frac{\frac{2}{1 + 2x}}{\sec^{2}x}). We have (\frac{\frac{2}{1+2\times0}}{\sec^{2}0}=\frac{2}{1}=2) Since (\lim_{x\rightarrow0}\ln y = 2), and (y = e^{\ln y}), then (\lim_{x\rightarrow0}y=e^{2})

Answer:

(e^{2})