Question

Найти предел функций:1)lim(x->∞) (x^4+3 x^2-5x)/(x^2+2)^2
2)lim(x->0) sin(2x^3)/x^3*sqrt(x+10)

Answer 1

$\lim_{x \to \infty} \frac{x^4+3x^2-5x}{(x^2+2)^2}= \lim_{x \to \infty} \frac{x^4+3x^2-5x}{x^4+4x^2+4}=\lim_{x \to \infty} \frac{\frac{x^4}{x^4}+\frac{3x^2}{x^4}-\frac{5x}{x^4}}{\frac{x^4}{x^4}+\frac{4x^2}{x^4}+\frac{4}{x^4}}=\frac{1+0-0}{1+0+0}=\\=\frac 11=1$

$\lim_{x \to 0} \frac{sin(2x^3)}{x^3*\sqrt{x+10}}= \lim_{x \to 0} \frac{2x^3}{x^3*\sqrt{x+10}}=\lim_{x \to 0} \frac{2}{\sqrt{x+10}}=\frac{2}{\sqrt{0+10}}=\frac{2}{\sqrt{10}}=\frac{2\sqrt{10}}{10}=\frac{\sqrt{10}}5$