Решите пределы! От души

1.
$\lim_{n \to \infty}\frac{(3n^2+5)(n-1)!}{(n+1)!}=\lim_{n \to \infty}\frac{3n^2+5}{n(n+1)}=\lim_{n \to \infty}\frac{3n^2+5}{n^2+n}=3$
2.
$\lim_{x\to\ 2}( \frac{1}{x-2}- \frac{12}{x^3-8})=\lim_{x\to\ 2} \frac{x^2+2x+4-12}{x^3-8}=\lim_{x\to\ 2} \frac{x^2+2x-8}{x^3-8}=$ $\lim_{x\to\ 2} \frac{(x-2)(x+4)}{x^3-8}=\lim_{x\to\ 2} \frac{x+4}{x^2+2x+4}= \frac{2+4}{2^2+2*2+4}= \frac{1}{2}$
3.
$\lim_{x \to \infty} ( \sqrt[3]{8x^3+3x^2+2}-2x)=$ $\lim_{x \to \infty} \frac{(\sqrt[3]{8x^3+3x^2+2}-2x)(\sqrt[3]{(8x^3+3x^2+2)^2}+2x*\sqrt[3]{8x^3+3x^2+2}+4x^2)}{\sqrt[3]{(8x^3+3x^2+2)^2}+2x*\sqrt[3]{8x^3+3x^2+2}+4x^2}=$ $\lim_{x \to \infty} \frac{8x^3+3x^2+2-8x^3}{\sqrt[3]{(8x^3+3x^2+2)^2}+2x*\sqrt[3]{8x^3+3x^2+2}+4x^2}=$ $\lim_{x \to \infty} \frac{3x^2+2}{\sqrt[3]{(8x^3+3x^2+2)^2}+2x*\sqrt[3]{8x^3+3x^2+2}+4x^2}= \frac{3}{4+4+4}= \frac{1}{4}$
4.
$\lim_{x \to 0} \frac{sin3x-sin7x}{sin5x}=\lim_{x \to 0}\frac{2*sin(-2x)*cos5x}{sin5x}=$ $\lim_{x \to 0}-\frac{4x}{5x}= - \frac{4}{5}$
5.
$\lim_{x \to 0} \frac{1-e^{6x}}{4x}=\lim_{x \to 0}-\frac{6x}{4x}=- \frac{3}{2}$