Question

Решите предел:
$\lim_{x \to0} \frac{2x^{2} }{1-cos5x}$

Answer 1

Воспользуемся таблицой бесконечно малых эквивалентных функций:

$\cos x \backsim 1 - \frac {x^2}2 \Rightarrow \cos 5x \backsim 1 - \frac{25x^2}2\\\lim\limits_{x\to0}\frac{2x^2}{1-\cos5x} = \lim\limits_{x\to0}\frac{2x^2}{1-1+\frac{25x^2}{2}} = \lim\limits_{x\to0}\frac{4x^2}{25x^2} = \frac4{25}$

Answer 2

$\displaystyle \tt \lim_{x \: \to \:0} \bigg(\frac{2x^2}{(1-cos(5x))}\bigg)= \lim_{x \: \to \: 0} \bigg(\frac{\frac{d}{d\:x}(2x^2)}{\frac{d}{d\:x}(1-cos(5x))}\bigg)= \lim_{x \: \to \: 0} \bigg(\frac{4x}{5\:sin(5x)}\bigg)=\\\\\\ \displaystyle \tt = \lim_{x \: \to \: 0} \bigg(\frac{\frac{d}{d\:x}(4x)}{\frac{d}{d\:x}(5\:sin(5x))}}\bigg)= \lim_{x \: \to \: 0} \bigg(\frac{4}{25\:cos(5x)}\bigg)=\frac{4}{25\:cos(5\cdot0)}=\\\\\\ \displaystyle \tt =\frac{4}{25}=\bold{0,16}$