Calculate $\lim_{x\to \infty} \left(\frac {\sqrt[n] 2 + \sqrt[n] 3} 2\right)^n$

Question

Can someone help me calculate the following limit:

$$\lim_{n\to \infty} \left(\frac {\sqrt[n] 2 + \sqrt[n] 3} 2\right)^n$$

For a start, how does $\sqrt[n]2-\sqrt[n]3$ behave as $n\to\infty$. — Angina Seng, Oct 29 '17 at 08:08
Taking logarithm of the limit and using the substitution $t=1/n$, you can then apply L'Hopital's rule. — Prasun Biswas, Oct 29 '17 at 08:25

score 1 · Accepted Answer · answered Oct 29 '17 at 08:40

$$\lim_{n\to \infty} \left(\frac {\sqrt[n]{2 }+ \sqrt[n] {3}} 2\right)^n =\lim_{n\to \infty} \left(1+\frac{x_n}{n}\right)^n$$

Where $$x_n=\frac{1}{2}(n(\sqrt[n]{2}-1)+n(\sqrt[n]{3}-1))$$

Now $$n(\sqrt[n]{a}-1)\to\ln a$$

So your limit is $$e^{\frac{1}{2}(\ln 2+\ln 3)}=\sqrt{6}$$

1 Answers1