Evaluating $ \lim_{x\to \infty} x \left({{\left(\frac{x}{x+1}\right)}^{x}-\frac{1}{e}}\right)$

Let $$y={\left(\frac{x}{x+1}\right)}^{x}\implies \log(y)=x \log\left(\frac{x}{x+1}\right)=-x \log\left(1+\frac 1 x\right)$$

Now, using Taylor expansion $$\log\left(1+\frac 1 x\right)=\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3 x^3}+O\left(\frac{1}{x^4}\right)$$ $$\log(y)=-1+\frac{1}{2 x}-\frac{1}{3 x^2}+O\left(\frac{1}{x^3}\right)$$ $$y=e^{\log(y)}=\frac{1}{e}+\frac{1}{2 e x}-\frac{5}{24 e x^2}+O\left(\frac{1}{x^3}\right)$$ $$x\left(y- \frac 1e\right)=\frac{1}{2 e}-\frac{5}{24 e x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and also how it is approached.

Use the above formula for $x=10$ (quite small). You should get $\frac{23}{48 e}\approx 0.176276$ while, using your pocket calculator, the result would be $\approx 0.176638$.