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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.

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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 $$ \frac{e^u-1}{u} \to 1, \quad \text{as}\quad u \to 0. $$

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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 $n$

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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.

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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$.

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