Proof $\lim\limits_{n \rightarrow \infty} \sqrt{n} (3^{\frac{1}{n}} - 2^{\frac{1}{n}}) = 0$ How would one go about proving the limit of sequence: 
$$\lim_{n \rightarrow \infty}\ \sqrt{n} (3^{\frac{1}{n}} - 2^{\frac{1}{n}}) = 0\,. $$
Epsilon definiton of limit seems to be too complicated and/or unsolvable (variable both in exponent and base). $\lim\limits_{x \rightarrow \infty} \ 3^{\frac{1}{n}}$ is clearly $1$, but how does it combine with $\sqrt{n}$? 
Is there an obvious upper bound that I'm missing? 
 A: From
$$1=3-2=\left(3^{\frac{1}{n}}\right)^n-\left(2^{\frac{1}{n}}\right)^n=\\
\left(3^{\frac{1}{n}}-2^{\frac{1}{n}}\right)\left( 3^{\frac{n-1}{n}}+3^{\frac{n-2}{n}} 2^{\frac{1}{n}}+3^{\frac{n-3}{n}} 2^{\frac{2}{n}}+...+3^{\frac{1}{n}} 2^{\frac{n-2}{n}} + 2^{\frac{n-1}{n}}\right)$$
we have
$$0<\sqrt{n}\left(3^{\frac{1}{n}}-2^{\frac{1}{n}}\right)=\frac{\sqrt{n}}{3^{\frac{n-1}{n}}+3^{\frac{n-2}{n}} 2^{\frac{1}{n}}+3^{\frac{n-3}{n}} 2^{\frac{2}{n}}+...+3^{\frac{1}{n}} 2^{\frac{n-2}{n}} + 2^{\frac{n-1}{n}}}<\\
\frac{\sqrt{n}}{2^{\frac{n-1}{n}}+2^{\frac{n-2}{n}} 2^{\frac{1}{n}}+2^{\frac{n-3}{n}} 2^{\frac{2}{n}}+...+2^{\frac{1}{n}} 2^{\frac{n-2}{n}} + 2^{\frac{n-1}{n}}}=
\frac{\sqrt{n}}{n2^{\frac{n-1}{n}}}<
\frac{1}{\sqrt{n}}$$
Or
$$0<\sqrt{n} \left(3^{\frac{1}{n}}-2^{\frac{1}{n}}\right)<\frac{1}{\sqrt{n}}$$
And use the squeeze theorem.
A: One possibility is evaluate $$\lim_{x\to 0}\dfrac{3^x-2^x}{\sqrt{x}}$$ using L'Hospital's Rule rule. Or otherwise we can use the identity $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + … + x y^{n-2} + y^{n-1})$ to obtain $$\sqrt{n} (3^{1/n} - 2^{1/n})=\dfrac{\sqrt{n}}{3^{(n-1)/n} + 3^{(n-2)/n} 2^{1/n} + … + 3^{1/n} 2^{(n-2)/n} + 2^{(n-1)/n}}.$$ Then $$\dfrac{3^{(n-1)/n} + 3^{(n-2)/n} 2^{1/n} + … + 3^{1/n} 2^{(n-2)/n} + 2^{(n-1)/n}}{n}\gt 2^{(n-1)/n}$$ as average is larger than the smallest term. Now use squeeze theorem. 
A: Simply use the definition of $a^x$ and Taylor's formula at order $1$: \begin{align}
\sqrt n\Bigl(3^{\tfrac1n}-^{\tfrac1n}\Bigr)&=\sqrt n\Bigl(\mathrm e^{\tfrac{\log 3}n}-\mathrm e^{\tfrac{\log2}n}\Bigr)=\sqrt n\biggl(1+\frac{\log 3}n+o\Bigl(\frac1n\Bigr)-1-\frac{\log 2}n-o\Bigl(\frac1n\Bigr)\biggr)\\
&=\frac{\log 3-\log2}{\sqrt n}+o\Bigl(\frac1{\sqrt n}\Bigr)\to 0.
\end{align}
A: $$\lim_{z\to 0}\frac{3^z-2^z}{z}=\log\frac{3}{2} $$
can be proved in a number of ways, for instance through De l'Hopital rule. It leads to
$$ 3^{1/n}-2^{1/n} \leq \frac{C}{n} \qquad (C>0)$$
as $n\to +\infty$, hence the given limit is zero by squeezing.
A: Consider $f(x) = x^{(1/n)}.$
$\dfrac{f(3) - f(2)}{1} = f'(t)$, $2 \lt t\lt3$
$ f'(t) = \dfrac{1}{nt^{(1-1/n)}}.$
Let $n \ge 2:$
$a_n:= √n(3^{(1/n)} -2^{(1/n)}) \lt  \dfrac{√n}{n2^{(1/2)}}$ 
$= \dfrac{√n}{n} =\dfrac{1}{√n}.$
Let $\epsilon >0$ be given.
There is a $n_0 \gt 1/\epsilon^2$. (Archimedes)
For $n\ge n_0 :$
$|a_n| \lt \dfrac{1}{√n} \le \dfrac{1}{√n_0} \lt \epsilon.$
