Evaluating $\lim\limits_{ n \to +\infty} (\sqrt{n}( e^{1/\sqrt{n}} -2^{1/\sqrt{n}}))^3$ 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?
 A: 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.
A: What Didier is suggesting the use of the little oh notation. This stands for the following:
$$e^x=1+x+o(x) \text{ for } x\to 0 $$ means that
$$\lim\limits_{x \to 0} \frac{e^x-1-x}{x} =0$$
or that we can approximate $e^x$ near $x=0$ very accurately by $1+x$, with an error that is very small in comparison to $x$ i.e $\dfrac{o(x)}{x}\to0$ (that is what $o(x)$ kinda means.)
To make things easier, put $\sqrt n=x$, to get
$$\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{e^x} - {2^x}}}{x}} \right)^3}$$
Now, we get the following:
$$\eqalign{
  & {e^x} = 1 + x + o\left( x \right)  \cr 
  & {2^x} = {e^{x\log 2}} = 1 + x\log 2 + o\left( x \right) \cr} $$
So this means:
$$\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + x + o\left( x \right) - 1 - x\log 2 - o\left( x \right)}}{x}} \right)^3}$$
$$\mathop {\lim }\limits_{x \to 0} {\left( {1 - \log 2 + \frac{{o\left( x \right)}}{x}} \right)^3}$$
But since we said $o(x)$ is an error that tends to zero in comparison to $x$, we get
$$\mathop {\lim }\limits_{x \to 0} {\left( {1 - \log 2 + \frac{{o\left( x \right)}}{x}} \right)^3} = {\left( {1 - \log 2} \right)^3}$$
Do you follow? Can you move on to the second case analogously?
A: This is more or less Brian M. Scott's answer, without converting to a limit at $0^+$. 
For the second limit, let's find the limit of the square. Using L'Hôpital's Rule twice:
$$\eqalign{
\lim_{n\rightarrow\infty} n^2(e^{1\over n}-e^{1\over n+1} )
&=\lim_{n\rightarrow\infty} {e^{1\over n}-e^{1\over n+1} \over 1/n^2}\cr
&=\lim_{n\rightarrow\infty} {(-1/n^2)e^{1/n}-(-1/(n+1))^2e^{1\over n+1} \over-2/n^3}\cr
&=\lim_{n\rightarrow\infty} { e^{1\over n}-({n\over n+1})^2e^{1\over n+1} \over2 /n}\cr
&=\lim_{n\rightarrow\infty} {(-1/n^2) e^{1\over n}-\Bigl[\,({n\over n+1})^2 \cdot{-1\over (n+1)^2}e^{1\over n+1}+{2n\over (n+1)^3}e^{1\over n+1}\Bigr] \over-2 /n^2}\cr
&=\lim_{n\rightarrow\infty} { e^{1\over n}-({n\over n+1})^4  e^{1\over n+1} 
+ {2n^3\over (n+1)^3}e^{1\over n+1} 
\over2 }\cr
&={1-1+2\over 2}\cr
&=1.
}
$$
So, the limit of the original expression is $\sqrt 1=1$.

For the first limit, I would initially ignore the cube, and first compute
$$
\lim_{n\rightarrow\infty} \bigl[\,{\sqrt n(e^{1/\sqrt n}- 2^{1/\sqrt n} )}\,\bigr]
=\lim_{n\rightarrow\infty}{ {  e^{1/\sqrt n}- e^{\ln 2/\sqrt n}  }\over 1/\sqrt n }.
$$
One application of  L'Hôpital's Rule will show the above limit is $1-\ln 2$; whence the original limit is $(1-\ln 2)^3$.
A: Because the limit of the cube of a sequence is the cube of the limit, it suffices to find and cube the following limit
$$\lim_{n \rightarrow \infty} \sqrt{n}(e^{1 \over \sqrt{n}} - 2^{1 \over \sqrt{n}})$$
Note that since $n$ going to infinity is the same as ${1 \over \sqrt{n}}$ going to zero, the above limit is
$$\lim_{x \rightarrow 0} {e^{x} - 2^{x} \over x}$$
L'hopital's rule gives that this limit is $1 - \ln2$, so the original limit is $(1 - \ln 2)^3$.
You can do something similar with the second one, but it's more complicated. The limit is the square root of
$$\lim_{n \rightarrow \infty} n^2(e^{1 \over n} - e^{1 \over n + 1})$$
Letting $x = {1 \over n}$ this becomes
$$\lim_{x \rightarrow 0} {e^x - e^{x \over x + 1} \over x^2}$$
Since $\lim_{x \rightarrow 0} e^x = 1$, you can factor out an $e^x$ in the above to obtain
$$\lim_{x \rightarrow 0} {1 - e^{-{x^2 \over x + 1}} \over x^2}$$
Since $x +1$ goes to $1$, this is the same as 
$$\lim_{x \rightarrow 0} {1 - e^{-{x^2 \over x + 1}} \over {x^2 \over x+1}}$$
So if you let $y = {x^2 \over x + 1}$ the above becomes 
$$\lim_{y \rightarrow 0} {1 - e^{-y} \over y}$$
By L'hopital's rule (or other methods) this limit is $1$, so the original limit is the square root of this, also $1$.
A: Let's solve them elementarily:
1st limit:
$$\lim_{ n \to +\infty}\left( \sqrt{n}\cdot \left( e^{\frac{1}{\sqrt{n}}} -2^{\frac{1}{\sqrt{n}}} \right) \right)^3=\lim_{ n \to +\infty}\left(\frac{e^{\frac{1}{\sqrt{n}}}-1}{\frac{1}{\sqrt{n}}} - \frac{2^{\frac{1}{\sqrt{n}}}-1}{\frac{1}{\sqrt{n}}}\right)^3=(1-\ln2)^{3}.$$
2nd limit:
$$L^2=\lim_{n \to +\infty} \left(n\cdot \sqrt{e^{\frac1n}-e^{\frac{1}{n+1}}}\right)^2=\lim_{n \to +\infty}n^2\left(\frac{e^{\frac{1}{n}}-1}{\frac{1}{n}}\cdot\frac{1}{n} - \frac{e^{\frac{1}{n+1}}-1}{\frac{1}{n+1}}\cdot\frac{1}{n+1}\right)$$
$$=\lim_{n \to +\infty} \frac{n^2}{n(n+1)}=1.$$
Therefore, $L=1$.
Remark: for the two limits i resorted to the auxiliary limit, $\lim_{x\to0} \frac{a^x-1}{x}=\log a$, $a>0$.
Q.E.D.
