## Calculus (3rd Edition)

$e^{-1}$.
To find the limit $$\lim _{x \rightarrow \infty}\left(\frac{x}{x+1}\right)^x$$ we have to find $$\ln \lim _{x \rightarrow \infty}\left(\frac{x}{x+1}\right)^x =\lim _{x \rightarrow \infty}\ln\left(\frac{x}{x+1}\right)^x\\ =\lim _{x \rightarrow \infty}x(\ln x-\ln(x+1) )=\lim _{x \rightarrow \infty}\frac{\ln x-\ln(x+1) }{1/x}=\frac{\infty}{0}$$ which is an intermediate form, so we can apply L’Hôpital’s Rule as follows $$\lim _{x \rightarrow \infty}\frac{\ln x-\ln(x+1) }{1/x}\\=\lim _{x \rightarrow \infty}\frac{(1/x) -1/(x+1) }{-1/x^2}=\lim _{x \rightarrow \infty}\frac{-x }{x+1}=\frac{\infty}{\infty}$$ Which is an intermediate form, so we can apply L’Hôpital’s Rule as follows $$\lim _{x \rightarrow \infty}\frac{-x }{x+1}=\lim _{x \rightarrow \infty}\frac{-1 }{1}=-1.$$ Now, $$\ln \lim _{x \rightarrow \infty}\left(\frac{x}{x+1}\right)^x=-1.\Longrightarrow \lim _{x \rightarrow \infty}\left(\frac{x}{x+1}\right)^x=e^{-1}.$$