## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to0^+}(1+x)^{\cot x}=e.$$
To solve this limit we will transform the expression under the limit. Since the natural exponential function and the natural logarithmic function are inverse to each other we have that $a=e^{\ln a}.$ This gives $(1+x)^{\cot x}=e^{\ln (1+x)^{\cot x}}=e^{\cot x\ln(1+x)}$ where we used the logarithmic rule $\ln b^a=a\ln b$. Also because the exponential function is continuous it can exchange places with the limit so we get $$L=\lim_{x\to0^+}(1+x)^{\cot x}=\lim_{x\to0^+}e^{\cot x\ln(1+x)}=e^{\lim_{x\to0^+}\cot x\ln(1+x)}=e^{\lim_{x\to0^+}\frac{\ln(1+x)}{\tan x}}=e^l,$$ where we have denoted $l=\lim_{x\to0^+}\frac{\ln(1+x)}{\tan x}.$ Let's solve this limit. "LR" will stand for "Apply L'Hopital's rule." $$l=\lim_{x\to0^+}\frac{\ln(1+x)}{\tan x}=\left[\frac{\ln(1+0)}{\tan 0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0^+}\frac{(\ln(1+x))'}{(\tan x)'}=\lim_{x\to0^+}\frac{\frac{1}{1+x}(1+x)'}{\frac{1}{\cos^2 x}}=\lim_{x\to0^+}\frac{\cos^2 x}{1+x}=\frac{\cos^2 0}{1+0}=1.$$ Putting this into the initial limit we get $$L=e^l=e^1=e.$$