Question

Решите пару предельчиков, пожалуйста:)

Answer 1

1)
$lim \frac{( {2}^{x} - 1)lncosx}{ {x}^{2} \times arcsin2x } = lim \frac{ {2}^{x} - 1 }{arcsin2x} \times lim \frac{lncosx}{ {x}^{2} }$
1.
$lim \frac{ {2}^{x} - 1}{arsin2x} = ( \frac{0}{0}) = lim\frac{ {2}^{x} ln2}{ \frac{2}{ \sqrt{1 - 4 {x}^{2} } } } = lim ({2}^{x - 1} ln2 \sqrt{1 - 4 {x}^{2} } ) = {2}^{0 - 1} ln2 \sqrt{1 - 4 \times 0} = \frac{ln2}{2}$
2.
$lim\frac{lncosx}{ {x}^{2} } = ( \frac{0}{0} ) = lim\frac{ \frac{1}{cosx} \times ( - sinx)}{2x} = - \frac{1}{2} lim \frac{1}{cosx} \times lim \frac{sinx}{x} = - \frac{1}{2} \times \frac{1}{cos0} \times lim \frac{cosx}{1} = - \frac{1}{2} \times 1 \times cos0 = - \frac{1}{2}$
$\frac{ln(2)}{2} \times ( - \frac{1}{2} ) = - \frac{ln(2)}{4}$
Ответ: - ln(2)/4.


2)
$lim(x - \pi)lnsinx = lim \frac{lnsinx}{ \frac{1}{x - \pi} } = lim \frac{ \frac{cosx}{sinx} }{ - \frac{1}{ {(x - \pi)}^{2} } } = - lim \frac{ {(x - \pi)}^{2} cosx}{sinx} = - limcosx \times lim \frac{ {(x - \pi)}^{2} }{sinx} = - cos\pi \times lim \frac{ 2(x - \pi) \times ( - 1)}{cosx} = - 1 \times ( - 1) \times ( - 1) \times lim\frac{2(x - \pi)}{cosx} = - \frac{2(\pi - \pi)}{cos\pi} = - \frac{2 \times 0}{ - 1} = 0$
Ответ: 0