Evaluate the following: $$ \lim_{x\to \infty} x \left({{\left(\frac{x}{x+1}\right)}^{x}-\frac{1}{e}}\right)$$
I tried to first solve the interior portion as $1^\infty$ indeterminate form but ended up getting a different indeterminate form of $\infty\cdot 0$.
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 $\approx 0.176638$.
$$\begin{align} \left(\frac{x}{x+1}\right)^x &=\left(1+\frac1x\right)^{-x}\\ &=\left(1+t\right)^{-1/t}\quad\left(t=\frac1x\right)\\ &=\exp{\left(-\frac{\ln{\left(1+t\right)}}{t}\right)}\\ &=\exp{\left(-\frac{t-t^2/2+o(t^2)}{t}\right)}\quad(\text{as }t\to0)\\ &=\exp{\left(-\left(1-\frac{t}2+o(t)\right)\right)}\\ &=\exp{\left(-1+\frac{t}2+o(t)\right)}\\ &=\frac1e\exp{\left(\frac{t}2+o(t)\right)}\\ &=\frac1e\left(1+\frac{t}2+o(t)\right)\quad(\text{as }t\to0)\\ &=\frac1e+\frac{t}{2e}+o(t)\\ \end{align}$$ Hence our limit becomes $$\begin{align} \lim_{x\to\infty}x\left(\left(\frac{x}{x+1}\right)^x-\frac1e\right) &=\lim_{t\to0^+}\frac1t\left(\frac1e+\frac{t}{2e}+o(t)-\frac1e\right)\\ &=\lim_{t\to0^+}\frac1t\left(\frac{t}{2e}+o(t)\right)\\ &=\lim_{t\to0^+}\left(\frac{1}{2e}+o(1)\right)\\ &=\frac1{2e}\\ \end{align}$$
