$$\lim_{x \to \infty} \bigg(\frac{x}{e} - x\Big(\frac{x}{x+1}\Big)^x\bigg)$$
I got the answer to the above limit by applying the L'Hôpital twice by first taking $x$ outside and then dividing it to make the limit of the form $\frac{0}{0}$
I want to to know whether there is any other method to solve this limit.
Any help would be appreciated.
Setting $u=\frac1x$ we can transform the expression to $$ \frac1u\left(\frac1e-(1+u)^{-\frac1u}\right)=\frac{\exp(-1)-\exp\left(-\frac1u\ln(1+u)\right)}{u}= \frac{\exp(-1)-\exp(-1+\frac1{2}u-\frac1{3}u^2+...)}{u}=... $$ using $\ln(1+u)=u-\frac12u^2+\frac13u^3-\frac14u^4+...$.
By the mean value theorem for $f(u)=\exp\left(-\frac1u\ln(1+u)\right)=\exp(-1+\frac1{2}u-\frac1{3}u^2+...)$ there exists some $0<h<u=\frac1x$ with $$ ...=-\frac{f(u)-f(0)}{u}=f'(h)=-(\tfrac12-\tfrac23h+...)\exp(-1+\tfrac1{2}h-\tfrac1{3}h^2+...) $$ so in the limit $x\to\infty$ we get $h\to 0$ and thus $-\frac1{2e}$ for the limit of the expression.
For large $x$,$$x\left(1+\frac1x\right)^{-x}=x\exp\left(-x\ln\left(1+\frac1x\right)\right)=x\exp\left(-1+\frac{1}{2x}-o\left(\frac{1}{x}\right)\right)=\frac{x}{e}+\frac{1}{2e}-o\left(\frac{1}{x}\right).$$Therefore, the sought limit is $\frac{-1}{2e}$, as expected.
