Question

pag 167 ex E4

lim x tinde la 0 ln(1+x^2) supra 5x^2

Answer 1

Observam ca atat numitorul cat si numaratorul tind catre 0, deci putem sa aplicam regula lui l'Hospital:
$\lim_{n \to \00} \frac{ln(1+x^2}{5x^2} = \lim_{n \to \00} \frac{(ln(1+x^2))'}{(5x^2)'}= \lim_{n \to \00} \frac{ \frac{2x}{x^2+1} }{10x}= \lim_{n \to \00} ( \frac{2x}{1+x^2} \frac{1}{10x} )$
$\lim_{n \to \00} (\frac{2x}{x^2+1} \frac{1}{10x} )= \lim_{n \to \00} \frac{1}{5(1+x^2)} = \frac{1}{5(0^2+1)} = \frac{1}{5} .$
Sper sa te ajute! :)