Find-lim-n-n-p-x-n-where-p-Z-

Question Number 8633 by Yozzias last updated on 19/Oct/16

Find \lim_{n \to \infty} \frac{n^{p}}{x^{n}} where p \in Z^{+} .

Commented by prakash jain last updated on 19/Oct/16

Does x cover entire range of R ?

For ∣ x ∣⩽ 1 limit is \infty .

For ∣ x ∣> 1 limit is 0 .

p ⩾ 1

Commented by Yozzias last updated on 19/Oct/16

Please consider the ∣ x ∣> 1 case! Thanks .

Answered by prakash jain last updated on 19/Oct/16

y = \frac{n^{p}}{x^{n}}

x > 1

x^{n} = e^{n \ln x}

x^{n} = \underset{i = 0}{\sum^{\infty}} \frac{{(n \ln x)}^{i}}{i!}

\frac{x^{n}}{n^{p}} = \underset{i = 0}{\sum^{p}} \frac{{(\ln x)}^{i}}{i! n^{p - i}} + \underset{i = p + 1}{\sum^{\infty}} \frac{n^{i - p} \ln x}{i!}

x > 1 \Rightarrow \ln x > 0

\lim_{n \to \infty} \frac{x^{n}}{n^{p}} = 0 + \infty = \infty

o r \lim_{n \to \infty} \frac{n^{p}}{x^{n}} = 0

f o r x < - 1 c a s e

o r \lim_{n \to \infty} \frac{n^{p}}{x^{n}} = \lim_{n \to \infty} \frac{1}{{(- 1)}^{n}} \frac{n^{p}}{y^{n}} = 0