sequences and series - Evaluating $limlimits_{ n to +infty} (sqrt{n}( e^{1/sqrt{n}} -2^{1/sqrt{n}}))^3$

I've got problems with calculating the limits in these two examples:

$$\begin{align*} &\lim_{ n \to +\infty}\left( \sqrt{n}\cdot \left( e^{\frac{1}{\sqrt{n}}} -2^{\frac{1}{\sqrt{n}}} \right) \right)^3\\ &\lim_{n \to +\infty} n\cdot \sqrt{e^{\frac1n}-e^{\frac{1}{n+1}}} \end{align*}$$

Can anybody help?

Answer

For the first one, let $u=\frac1{\sqrt n}$, and note that $u\to 0^+$ as $n\to\infty$. Then

$$\sqrt n\left(e^{\frac1{\sqrt n}}-2^{\frac1{\sqrt n}}\right)=\frac1u\left(e^u-2^u\right)=\frac{e^u-2^u}u\;,$$ so you’re interested in $$\lim_{u\to 0^+}\frac{(e^u-2^u)^3}{u^3}\;,$$ and I expect that you know a way to deal with that kind of limit.

I can make a similar trick work for the second one, but it gets a bit messier. First, I’m actually going to look at

$$\lim_{n\to\infty}n^2\left(e^{\frac1n}-e^{\frac1{n+1}}\right)$$ and then take its square root to get the desired limit.

Let $u=\frac1n$, so that $n=\frac1u$, $n+1=\frac1u+1=\frac{u+1}u$, and $\frac1{n+1}=\frac{u}{u+1}=1-\frac1{u+1}$. Again $u\to 0^+$ as $n\to\infty$, so I look at

$$\lim_{u\to 0^+}\frac{e^u-e^{1-\frac1{u+1}}}{u^2}\;;$$ applying l’Hospital’s rule takes a little more work this time, but it’s still eminently feasible.