Find-lim-x-x-2-ln-1-1-x-x-without-DL-and-if-you-have-to-use-the-hospital-rule-please-justify-that-your-function-is-C-1-

Question Number 79689 by ~blr237~ last updated on 27/Jan/20

F i n d \lim_{x \to \infty} x^{2} l n (1 + \frac{1}{x}) - x w i t h o u t D L a n d i f y o u h a v e t o u s e t h e

h o s p i t a l r u l e p l e a s e j u s t i f y t h a t y o u r f u n c t i o n i s C^{1}

Commented by mr W last updated on 27/Jan/20

\lim_{x \to \infty} x^{2} l n (1 + \frac{1}{x}) - x

= \lim_{x \to 0} \frac{1}{x^{2}} l n (1 + x) - \frac{1}{x}

= \lim_{x \to 0} \frac{1}{x^{2}} (x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \dots) - \frac{1}{x}

= \lim_{x \to 0} (- \frac{1}{2} + \frac{x}{3} - \frac{x^{2}}{4} + \dots)

= - \frac{1}{2}

Commented by john santu last updated on 27/Jan/20

{mac}^{'} laurin series