Question

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

Answer 1

Воспользуемся таблицой бесконечно малых эквивалентных функций:

$cos x backsim 1 - frac {x^2}2 Rightarrow cos 5x backsim 1 - frac{25x^2}2\limlimits_{xto0}frac{2x^2}{1-cos5x} = limlimits_{xto0}frac{2x^2}{1-1+frac{25x^2}{2}} = limlimits_{xto0}frac{4x^2}{25x^2} = frac4{25}$

Answer 2

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