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: \(e^{2}\)