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.
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