<s>[INST] 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? 
 [/INST] 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.
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