Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{\ln x}=1.$$
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 $\left(1+\frac{1}{x}\right)^{\ln x}=e^{\ln \left(1+\frac{1}{x}\right)^{\ln x}}=e^{\ln x\ln\left(1+\frac{1}{x}\right)}$ 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 $$L=\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^{\ln x}=\lim_{x\to\infty}e^{\ln x\ln\left(1+\frac{1}{x}\right)}=e^{\lim_{x\to\infty}\ln x\ln\left(1+\frac{1}{x}\right)}=e^l,$$ where we have denoted $l=\lim_{x\to\infty}\ln x\ln\left(1+\frac{1}{x}\right).$ Now let's calculate this limit. "LR" will stand for "Apply L'Hopital's rule". $$l=\lim_{x\to\infty}\ln x\ln\left(1+\frac{1}{x}\right)=\lim_{x\to\infty}\frac{\ln\left(1+\frac{1}{x}\right)}{\frac{1}{\ln x}}=\left[\frac{\ln(1+1/\infty)}{\frac{1}{\ln\infty}}\right]=\left[\frac{\ln (1+0)}{\frac{1}{\infty}}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to\infty}\frac{\left(\ln\left(1+\frac{1}{x}\right)\right)'}{\left(\frac{1}{\ln x}\right)'}=\lim_{x\to\infty}\frac{\frac{1}{1+\frac{1}{x}}\left(1+\frac{1}{x}\right)'}{-\frac{1}{\ln^2x}(\ln x)'}=\lim_{x\to\infty}\frac{\frac{x}{1+x}\left(-\frac{1}{x^2}\right)}{-\frac{1}{\ln^2 x}\frac{1}{x}}=\lim_{x\to\infty}\frac{\ln^2 x}{1+x}=\left[\frac{\ln^2\infty}{1+\infty}\right]=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(\ln^2x)'}{(1+x)'}=\lim_{x\to\infty}\frac{2\ln x(\ln x)'}{1}=\lim_{x\to\infty}\frac{2\ln x}{x}=\left[\frac{2\ln\infty}{\infty}\right]=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to\infty}\frac{(2\ln x)'}{(x)'}=\lim_{x\to\infty}\frac{2\frac{1}{x}}{1}=\lim_{x\to\infty}\frac{2}{x}=\left[\frac{2}{\infty}\right]=0.$$ Putting this into the initial limit we have $$L=e^l=e^0=1.$$