Question

lim (где х->бесконечности)3х(в стп.2)-2х+1/2х(в стп.2)+х-3

Answer 1

$<var>\lim_{n \to \infty} \frac{3x^2 -2x +1}{2x^2+x-3} = \lim_{n \to \infty} \frac{\frac{3x^2}{x^2} -\frac{2x }{x^2}+\frac{1}{x^2}}{\frac{2x^2}{x^2}+\frac{x}{x^2}-\frac{3}{x^2}} = \lim_{n \to \infty} \frac{3-0 +0}{2+0-0}=</var>$ $<var>=\frac{3}{2}=1.5</var>$

Answer 2

 

$<var>\lim_{x \to \infty} \frac{3x^2-2x+1}{2x^2+x-3}=\lim_{x \to \infty} \frac{x^2(3-\frac{2}{x}+\frac{1}{x^2})}{x^2(2+\frac{1}{x}-\frac{3}{x^2})}=\lim_{x \to \infty} \frac{3-\frac{2}{x}+\frac{1}{x^2}}{2+\frac{1}{x}-\frac{3}{x^2}}= \\ =\frac{3-0+0}{2+0-0}=1,5</var>$