Вычеслить пределы (полностью с решением)

Ответ

Автор - MrSolution

Ответ:

$a) dfrac{51}{14};; ; b)dfrac{2}{5}$

Пошаговое объяснение:

$lim_{x to 7} dfrac{x^3-7x^2+2x-14}{x^2-49}= lim_{x to 7} dfrac{x^3+2x-7x^2-14}{x^2-49}= lim_{x to 7} dfrac{x(x^2+2)-7(x^2+2)}{(x+7)(x-7)}=lim_{x to 7} dfrac{(x^2+2)(x-7)}{(x+7)(x-7)}=lim_{x to 7} dfrac{x^2+2}{x+7}=dfrac{7^2+2}{7+7}=dfrac{51}{14}$

$lim_{x to 0}cot5xtimes tan 2x= lim_{x to 0}dfrac{cos 5x times sin 2x}{sin 5x times cos2x}=lim_{x to 0}dfrac{dfrac{d}{dx}(cos 5x times sin 2x)}{dfrac{d}{dx}(sin 5x times cos2x)}=lim_{x to 0} dfrac{-5sin 5xtimes sin 2x + 2cos 5xtimes cos 2x}{5cos 5x times cos 2x - 2sin 5x times sin 2x}=$ $dfrac{-5sin 5times 0times sin 2times 0 + 2cos 5times 0times cos 2times 0}{5cos 5times 0 times cos 2times 0 - 2sin 5times 0 times sin 2times 0}=dfrac{2}{5}$