1. Tìm lim [{căn(x^2+x+2) – căn(3-x)} / {x^4+x}]
x–>(-1)-
2. Tìm lim [{5} / {4x^3-x^3+2}]
x–>-vô cực
1. Tìm lim [{căn(x^2+x+2) – căn(3-x)} / {x^4+x}]
x–>(-1)-
2. Tìm lim [{5} / {4x^3-x^3+2}]
x–>-vô cực
1.
Ta có:
Tử $\to \sqrt{1-1+2}-\sqrt{3+1}=\sqrt2-2<0$
Mẫu $\to 0$
$x\to (-1)^-\Rightarrow x<-1\Rightarrow x^3<-1\Rightarrow x(x^3+1)>0$
$\to$ $\lim\limits_{x\to (-1)^-}\dfrac{ \sqrt{x^2+x+2}-\sqrt{3-x} }{x^4+x}=-\infty$
2.
$\lim\limits_{x\to -\infty}\dfrac{5}{4x^3-x^3+2}$
$=\lim\limits_{x\to -\infty}\dfrac{ \dfrac{5}{x^3}}{3+\dfrac{2}{x^3}}$
$=0$
Giải thích các bước giải:
1.Ta có:
$\lim_{x\to(-1)^-}\dfrac{\sqrt{x^2+x+2}-\sqrt{3-x}}{x^4+x}$
$=\lim_{x\to(-1)^-}\dfrac{\sqrt{x^2+x+2}-\sqrt{3-x}}{x(x^3+1)}$
$=-\dfrac{\sqrt{(-1)^2+(-1)+2}-\sqrt{3-(-1)}}{(-1)((-1)^3+1)}$ vì $x\to (-1)^-\to x\le -1\to x^3\le -1\to x^3+1\le 0$
$=+\infty$
2.Ta có:
$\lim_{x\to-\infty}\dfrac{5}{4x^3-x^3+2}$
$=\lim_{x\to-\infty}\dfrac{5}{3x^3+2}$
$=0$