Calculate limit without de l'Hospital rule

Question

I have problem with showing that $\displaystyle \lim_{n \to \infty} n (\frac{e}{(1+\frac{1}{n})^n}-1)=\frac{1}{2} $ without de l'Hospital rule

I thought to use the rule $\displaystyle \lim_{n\to \infty} n(\ln a-1)=a$

Answer 1

If $f(x)$ is a convex function on the interval $[a,a+1]$ then: $$f\left(a+\frac{1}{2}\right)\leq\int_{a}^{a+1}f(x)\,dx\leq\frac{1}{2}\left(f(a)+f(a+1)\right),$$ hence: $$\frac{2}{2n+1}\leq\log\left(1+\frac{1}{n}\right)=\int_{n}^{n+1}\frac{dt}{t}\leq \frac{1}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right)\tag{1}$$ and: $$ \frac{1}{2n+2}\leq 1-n\log\left(1+\frac{1}{n}\right)\leq\frac{1}{2n+1}\tag{2}$$ so the value of our limit is between: $$\lim_{n\to +\infty} n\left(e^{\frac{1}{2n+2}}-1\right) $$ and $$\lim_{n\to +\infty} n\left(e^{\frac{1}{2n+1}}-1\right) $$ so the limit is $\frac{1}{2}$ by squeezing.