Помогите с пределами, пожалуйста

Решите задачу:

$1) \lim_{x \to \infty} { (\sqrt[3]{x+1} -\sqrt[3]{x})} =\lim_{x \to \infty} { \frac{\sqrt[3]{x+1}^3 -\sqrt[3]{x}^3}{\sqrt[3]{(x+1)^2}+\sqrt[3]{(x+1)x} +\sqrt[3]{x^2}}}=$
$\lim_{x \to \infty} { \frac{x+1-x}{\sqrt[3]{(x+1)^2}+\sqrt[3]{(x+1)x} +\sqrt[3]{x^2}}}=\lim_{x \to \infty} { \frac{1}{\sqrt[3]{(x+1)^2}+\sqrt[3]{(x+1)x} +\sqrt[3]{x^2}}}=0$
$2) \lim_{x \to 0} { \frac{1-cos7x}{x^2} } =\lim_{x \to 0} { \frac{1-(1-2sin^2(3,5x))}{ \frac{1}{3,5^2} (3,5x)^2} } =24,5*\lim_{x \to 0} { \frac{sin^2(3,5x))}{ (3,5x)^2} }$
$24,5*\lim_{x \to 0} { (\frac{sin(3,5x)}{ 3,5x})^2 } =24,5*1^2=24,5$
$3) \lim_{x \to \infty} {(\frac{x+8}{x-2})^x} = \lim_{x \to \infty} {(1+\frac{6}{x-2})^x}; t=\frac{x-2}{6} ;x=6t+2$
$\lim_{x \to \infty} {(1+\frac{6}{x-2})^x}=\lim_{t \to \infty} {(1+\frac{1}{t})^{6t+2}}=$
$=\lim_{t \to \infty} {((1+\frac{1}{t})^2*((1+\frac{1}{t})^{t})^6)}=1*e^6=e^6$