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


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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

$\lim_{x \to \infty} x^{2} l n (1 + \frac{1}{x}) - x$ $= \lim_{x \to 0} \frac{1}{x^{2}} l n (1 + x) - \frac{1}{x}$ $= \lim_{x \to 0} \frac{1}{x^{2}} (x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + . . .) - \frac{1}{x}$ $= \lim_{x \to 0} (- \frac{1}{2} + \frac{x}{3} - \frac{x^{2}}{4} + . . .)$ $= - \frac{1}{2}$
Commented by john santu last updated on 27/Jan/20

${mac}^{'} laurin series$
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