lim-x-ln-x-x-You-can-only-use-series-expansion-sandwich-theorem-

Question Number 34843 by rahul 19 last updated on 11/May/18

\underset{x \to \infty}{\lim_{}} \frac{\ln x}{x} = ?

Y o u c a n o n l y u s e

s e r i e s e x p a n s i o n / s a n d w i c h t h e o r e m!

Commented by abdo mathsup 649 cc last updated on 11/May/18

c h a n g e m e n t x = 1 + \frac{1}{t} g i v e

\frac{l n x}{x} = \frac{l n (1 + \frac{1}{t})}{1 + \frac{1}{t}} = \frac{l n (1 + t) - l n (t)}{\frac{t + 1}{t}}

= \frac{t}{t + 1} {l n (1 + t) - l n (t)} = \frac{t l n (1 + t) - t l n (t)}{t + 1}

= \frac{t}{t + 1} l n (1 + t) - \frac{t l n (t)}{t + 1} b u t

l n {(1 + t)}^{.} = \frac{1}{1 + t} = \sum_{n = 0}^{\infty} {(- 1)}^{n} t^{n}

l n (1 + t) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} t^{n + 1} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} t^{n}

\frac{l n (x)}{x} = \frac{t}{t + 1} (t - \frac{t^{2}}{2} + \frac{t^{3}}{3} + \dots) - t l n (t) \sum_{n = 0}^{\infty} {(- 1)}^{n} t^{n}

= \frac{t^{2}}{t + 1} (1 - \frac{t}{2} + \frac{t^{2}}{3} + \dots) - \sum_{n = 0}^{\infty} {(- 1)}^{n} t^{n + 1} l n (t) \to 0

w h e n t \to 0 s o {l i m}_{x \to + \infty} \frac{l n x}{x} = 0