calculate-lim-n-n-2-n-

Question Number 8066 by Chantria last updated on 29/Sep/16

c a l c u l a t e

\underset{n \to \infty}{l i m} \frac{n}{2^{\sqrt{n}}}

Answered by prakash jain last updated on 02/Oct/16

2^{\sqrt{n}} = e^{\sqrt{n} \ln 2} = 1 + \sqrt{n} \ln 2 + \frac{{(\sqrt{n} \ln 2)}^{2}}{2!} + \frac{{(\sqrt{n} \ln 2)}^{3}}{3!} + . .

\frac{n}{2^{\sqrt{n}}} = \frac{n}{1 + \sqrt{n} \ln 2 + \frac{{(\sqrt{n} \ln 2)}^{2}}{2!} + \frac{{(\sqrt{n} \ln 2)}^{3}}{3!} + . .}

= \frac{\frac{n}{n}}{1 + \frac{\ln \sqrt{2}}{n} + \frac{\ln 2}{2!} + \frac{\sqrt{n} \ln 2}{3!} + (higher power of n)}

= \frac{1}{1 + \frac{\ln \sqrt{2}}{n} + \frac{\ln 2}{2!} + \frac{\sqrt{n} \ln 2}{3!} + (higher power of n)}

\lim_{n \to \infty} \frac{n}{2^{\sqrt{n}}} = \frac{1}{1 + 0 + \frac{\ln 2}{2!} + \infty} = 0