Zadacha Kuznecov Predely 10-7

Условие задачи

Вычислить предел функции:

$\lim _{x\to 8}{\frac {{\sqrt {9+2x}}-5}{{\sqrt[{3}]{x}}-2}}$

Решение

$\lim _{x\to 8}{\frac {{\sqrt {9+2x}}-5}{{\sqrt[{3}]{x}}-2}}=\left\{{\frac {0}{0}}\right\}=\lim _{x\to 8}{\frac {\left({\sqrt {9+2x}}-5\right)\left({\sqrt {9+2x}}+5\right)}{\left({\sqrt[{3}]{x}}-2\right)\left({\sqrt {9+2x}}+5\right)}}=$

$=\lim _{x\to 8}{\frac {9+2x-25}{\left({\sqrt[{3}]{x}}-2\right)\left({\sqrt {9+2x}}+5\right)}}=\lim _{x\to 8}{\frac {2x-16}{\left({\sqrt[{3}]{x}}-2\right)\left({\sqrt {9+2x}}+5\right)}}=$

$=\lim _{x\to 8}{\frac {2(x-8)}{\left({\sqrt[{3}]{x}}-2\right)\left({\sqrt {9+2x}}+5\right)}}=\lim _{x\to 8}{\frac {2\left({\sqrt[{3}]{x}}-2\right)\left({\sqrt[{3}]{x^{2}}}+2{\sqrt[{3}]{x}}+4\right)}{\left({\sqrt[{3}]{x}}-2\right)\left({\sqrt {9+2x}}+5\right)}}=$

$=\lim _{x\to 8}{\frac {2\left({\sqrt[{3}]{x^{2}}}+2{\sqrt[{3}]{x}}+4\right)}{{\sqrt {9+2x}}+5}}={\frac {2\left({\sqrt[{3}]{8^{2}}}+2{\sqrt[{3}]{8}}+4\right)}{{\sqrt {9+2\cdot 8}}+5}}=$

$={\frac {2\left(2^{2}+2\cdot 2+4\right)}{{\sqrt {25}}+5}}={\frac {2\left(4+4+4\right)}{5+5}}={\frac {2\cdot 12}{10}}={\frac {12}{5}}=2{\frac {2}{5}}$