Вычислить пределы функций

Решите задачу:

$\displaystyle \lim_{x \to \8} \frac{\sqrt{9+2x}-5}{\sqrt[3]{x}-2}=|*\frac{(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)(\sqrt{9+2x}+5)}{(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)(\sqrt{9+2x}+5)}|=\\\\ \lim_{x \to \8} \frac{((\sqrt{9+2x})^2-25)*(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}{(\sqrt[3]{x^3}-8)(\sqrt{9+2x}+5)}=\\\\= \lim_{x \to \8}\frac{(2x-16)(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}{(x-8)(\sqrt{9+2x}+5)}=\\\\= \lim_{x \to \8} \frac{2(\sqrt[3]{x^2}+2\sqrt[3]{x}+4)}{\sqrt{9+2x}+5}=\frac{2(\sqrt[3]{8^2}+2\sqrt[3]{8}+4)}{(\sqrt{9+2*8}+5)}=$

$\displaystyle =\frac{2*(4+4+4)}{5+5}=\frac{2*12}{10}=\frac{12}{5}$