1. Tìm lim [{-3} / {căn(5x^2+x+2) + 4x}]
x–>-vô cực
2. Tìm lim [{(x-2)^3 + 8} / {x}]
x–>0
1. Tìm lim [{-3} / {căn(5x^2+x+2) + 4x}]
x–>-vô cực
2. Tìm lim [{(x-2)^3 + 8} / {x}]
x–>0
Giải thích các bước giải:
1.Ta có:
$\lim_{x\to -\infty}\dfrac{-3}{\sqrt{5x^2+x+2}+4x}$
$=\lim_{x\to -\infty}\dfrac{-3}{(-x)(\sqrt{5-\dfrac1x+\dfrac2{x^2}}-4)}$
$=\dfrac{-3}{+\infty(\sqrt{5-0+0}-4)}$
$=0$
2.Ta có:
$\lim_{x\to0}\dfrac{(x-2)^3+8}{x}$
$=\lim_{x\to0}\dfrac{(x-2)^3+2^3}{x}$
$=\lim_{x\to0}\dfrac{(x-2+2)((x-2)^2-2(x-2)+2^2)}{x}$
$=\lim_{x\to0}\dfrac{x((x-2)^2-2(x-2)+2^2)}{x}$
$=\lim_{x\to0}(x-2)^2-2(x-2)+2^2$
$=(0-2)^2-2(0-2)+2^2$
$=12$
1. $\lim\limits_{x\to -\infty}\dfrac{ -3}{\sqrt{5x^2+x+2}+4x}$ $=\lim\limits_{x\to -\infty}\dfrac{ \dfrac{-3}{x} }{ -\sqrt{5+\dfrac{1}{x}+\dfrac{2}{x^2}}+4}$ $=0$ 2. $\lim\limits_{x\to 0}\dfrac{(x-2)^3+2^3}{x}$ $=\lim\limits_{x\to 0}\dfrac{x[(x-2)^2-2(x-2)+4 ]}{x}$ $=\lim\limits_{x\to 0}(x-2)^2-2(x-2)+4$ $=2^2+4+4=12$