Computing a limit of root difference How do you compute the following limit by only using known limits, basic limit properties and theorems, such as the squeeze theorem. 
$\lim\limits_{n\to\infty} \sqrt[]{n}·(\sqrt[n]{3}-\sqrt[n]{2})$
Everything I´ve tried led me to undefined expressions and when I tried to bound the the expression in brackets I couldn't figure out proper lower bound..
The upper bound is easy, we can take $\sqrt[n]{3}$ which tends to $1$ but I can't think of any lower bound that wouldn't tend to $0$. 
Thanks.
 A: 
How do you compute the following limit by only using known limits, basic limit properties and theorems, such as the squeeze theorem. 

One may write, as $n \to \infty$,
$$
\sqrt[]{n}·(\sqrt[n]{3}-\sqrt[n]{2})=\frac{\log 3}{\sqrt{n}}\left(\frac{e^{\large \frac{\log 3}{n}}-1}{\frac{\log 3}{n}} \right)-\frac{\log 2}{\sqrt{n}}\left(\frac{e^{\large \frac{\log 2}{n}}-1}{\frac{\log 2}{n}} \right) \to 0
$$ using the known limit
$$
\frac{e^u-1}{u} \to 1, \quad \text{as}\quad u \to 0.
$$
A: How about $\sqrt[n]3=e^{\frac 1n \log 3}$, so $\sqrt[]{n}·(\sqrt[n]{3}-\sqrt[n]{2}) \approx \sqrt n(1+\frac 1n\log 3-1-\frac 1n \log 2)\approx (\log 3-\log 2)n^{-1/2}\to 0$  You have to justify neglecting the higher order terms, but they will have higher negative exponents on $n$
A: We will show, for $a > 1$, $\lim_n \to \infty \sqrt{n}(a^{\frac{1}{n}} - 1) = 0$
This If this holds true, we can take the two limits for $а=3$ and $а=2$ and subtract them tо get the answer 0. 
If you are allowed to use derivatives, then for $f(x) = a^x$, the definition of derivative gives
$$ \lim_{n \to \infty} n(a^{\frac{1}{n}} - 1)= \lim_{n \to \infty} \frac{a^{\frac{1}{n}} - 1}{\frac{1}{n}}=f'(0) = \ln(a)$$
 $\frac{1}{\sqrt{n} \to 0}$, so multiplying the above limits gives
$\lim_n \to \infty \sqrt{n}(a^{\frac{1}{n}} - 1) = 0$
which we wanted to prove. 
If you don't know derivatives, you can use  Bernoulli's inequality
$(1 + x)^r \leq 1 + rx$ for $x \geq 0,  0 \leq r \leq 1$
$$0 \leq \sqrt{n}(a^{\frac{1}{n}} - 1 )\leq \sqrt{n}( 1 + \frac{a-1}{n} - 1) \to 0$$ and the result follows by the Sandwich theorem.
A: If
$a > b$,
$\begin{array}\\
\sqrt[n]{a}-\sqrt[n]{b}
&=(\sqrt[n]{a}-\sqrt[n]{b})\frac{\sum_{k=0}^{n-1}a^{k/n}b^{(n-1-k)/n}}{\sum_{k=0}^{n-1}a^{k/n}b^{(n-1-k)/n}}\\
&=\frac{a-b}{\sum_{k=0}^{n-1}a^{k/n}b^{(n-1-k)/n}}\\
\end{array}
$
Since
$\sum_{k=0}^{n-1}a^{k/n}b^{(n-1-k)/n}
>\sum_{k=0}^{n-1}b^{k/n}b^{(n-1-k)/n}
=nb^{(n-1)/n}
$
and,
similarly,
$\sum_{k=0}^{n-1}a^{k/n}b^{(n-1-k)/n}
<na^{(n-1)/n}
$,
$\frac{a-b}{na^{(n-1)/n}}
< \sqrt[n]{a}-\sqrt[n]{b}
<\frac{a-b}{nb^{(n-1)/n}}
$.
Therefore
$\sqrt{n}\frac{a-b}{na^{(n-1)/n}}
< \sqrt{n}(\sqrt[n]{a}-\sqrt[n]{b})
<\sqrt{n}\frac{a-b}{nb^{(n-1)/n}}
$
or
$\frac{a-b}{\sqrt{n}a^{(n-1)/n}}
< \sqrt{n}(\sqrt[n]{a}-\sqrt[n]{b})
<\frac{a-b}{\sqrt{n}b^{(n-1)/n}}
$.
From the upper bound,
we see that
$\sqrt{n}(\sqrt[n]{a}-\sqrt[n]{b})
$ converges to zero
like
$\frac1{\sqrt{n}}
$.
More generally,
this shows that
$n^u(\sqrt[n]{a}-\sqrt[n]{b})
\to 0$
like
$n^{u-1}$
if
$0 < u < 1$,
$n(\sqrt[n]{a}-\sqrt[n]{b})
$
is finite,
and
$n^u(\sqrt[n]{a}-\sqrt[n]{b})
\to \infty$
like
$n^{u-1}$
if
$ u > 1$.
