A square root limit

Let $c>0$ . Prove that

$\lim _{n\rightarrow +\infty} \left ( \frac {1+ \sqrt [n] {c}}{2} \right ) ^n= \sqrt {c}$

Solution

It holds that

$x^{1/n} = 1 + \frac{\log x}{n} + \mathcal{O} \left ( \frac{1}{n^2} \right )$

Thus,

$\begin{align*} \lim_{n \rightarrow +\infty} \left ( \frac{1+\sqrt[n]{c}}{2} \right )^n &= \lim_{n \rightarrow +\infty} \left ( \frac{1+c^{1/n}}{2} \right )^n \\ &=\lim_{n \rightarrow +\infty} \left ( \frac{1}{2} +\frac{1}{2}\left ( 1+ \frac{\log c}{n} + \mathcal{O}\left ( \frac{1}{n^2} \right ) \right ) \right )^n \\ &= \lim_{n \rightarrow +\infty} \left ( 1 + \frac{\log c}{2n} + \frac{1}{2} \mathcal{O}\left ( \frac{1}{n^2} \right ) \right )^n \\ &= \lim_{n \rightarrow +\infty} \left ( 1+ \frac{\log \sqrt{c}}{n} + \frac{1}{2} \mathcal{O}\left ( \frac{1}{n^2} \right ) \right )^n \\ &= \exp \left ( \log \sqrt{c} \right ) \\ &= \sqrt{c} \end{align*}$

Note: Similarly, it holds that $\lim \limits_{n \rightarrow +\infty} \left ( 2\sqrt[n]{x}-1 \right )^n =x^2$ .

A square root limit

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