Answer
$ e^{-1}$.
Work Step by Step
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, $$\ln \lim _{x \rightarrow \infty}\left(\frac{x}{x+1}\right)^x=-1.\Longrightarrow \lim _{x \rightarrow \infty}\left(\frac{x}{x+1}\right)^x=e^{-1}. $$
