Пожайлуста решите задания с пределами

Решите задачу:

$\lim_{x \to 2} \frac{x^2-4}{x-2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} x+2==4$
$\lim_{x \to 1} \frac{x^2-3x+2}{x^2+x-2} = \lim_{x \to 1} \frac{(x-1)(x-2)}{(x-1)(x+2)} = \lim_{x \to 1} \frac{x-2}{x+2}= \frac{1-2}{1+2} = -\frac{1}{3}$
$\lim_{x \to 2} \frac{x^2-7x+10}{x^3-8} = \lim_{x \to 2} \frac{(x-2)(x-5)}{(x-2)(x^2+2x+4)} = \lim_{x \to 2} \frac{x-5}{x^2+2x+4} = \\ \frac{2-5}{2^2+2*2+4} = -\frac{3}{12} =- \frac{1}{4}$

$\lim_{x \to 0} \frac{sin2x}{4x} = \frac{1}{2} \lim_{x \to 0} \frac{sin2x}{2x} = \frac{1}{2}$
$\lim_{x \to 0} \frac{sin12x}{tg4x} = \lim_{x \to 0} \frac{ 12x\frac{sin12x}{12x} }{ 4x\frac{tg4x}{4x} } = \frac{12}{4} =3$
$\lim_{x \to \infty} (1+ \frac{1}{2x} ) ^x= \lim_{x \to \infty} ( (1+ \frac{1}{2x} ) ^{2x})^{1/2}=e^{1/2}= \sqrt{e}$
$\lim_{x \to 0} (1-2x)^{1/x}= \lim_{x \to 0}( (1+(-2x))^{-1/2x})^{-2}=e^{-2}= \frac{1}{e^2}$