Bài tập nâng cao · Bài 8
Bài tập nâng cao Giới hạn của hàm số · Bài 8
Tính giới hạn
a. $\lim\limits_{x\to +\infty}\frac{2x^{2}-3x+1}{-2+3x-4x^{2}}$
b. $\lim\limits_{x\to -\infty}\frac{\sqrt{x^{2}+2x}+3x}{\sqrt{4x^{2}+1}-x+3}$
a. $\lim\limits_{x\to +\infty}\frac{2x^{2}-3x+1}{-2+3x-4x^{2}}$
b. $\lim\limits_{x\to -\infty}\frac{\sqrt{x^{2}+2x}+3x}{\sqrt{4x^{2}+1}-x+3}$
Xem lời giải
Lời giải
a. $\lim\limits_{x\to +\infty}\frac{2x^{2}-3x+1}{-2+3x-4x^{2}}=\lim\limits_{x\to +\infty}\frac{2-\frac{3}{x}+\frac{1}{x^{2}}}{\frac{-2}{x^{2}}+\frac{3}{x}-4}=\frac{-1}{2}$
b. $\lim\limits_{x\to -\infty}\frac{\sqrt{x^{2}+2x}+3x}{\sqrt{4x^{2}+1}-x+3}=\lim\limits_{x\to -\infty}\frac{\sqrt{x^{2}.\left( 1+\frac{2}{x} \right)}+3x}{\sqrt{x^{2}.\left( 4+\frac{1}{x^{2}} \right)}-x+3}$
$=\lim\limits_{x\to -\infty}\frac{\left| x \right|\sqrt{1+\frac{2}{x}}+3x}{\left| x \right|\sqrt{4+\frac{1}{x^{2}}}-x+3}=\lim\limits_{x\to -\infty}\frac{-x.\sqrt{1+\frac{2}{x}}+3x}{-x.\sqrt{4+\frac{1}{x^{2}}}-x+3}=\lim\limits_{x\to -\infty}\frac{-\sqrt{1+\frac{2}{x}}+3}{-\sqrt{4+\frac{1}{x^{2}}}-1+\frac{3}{x}}=\frac{-2}{3}$
b. $\lim\limits_{x\to -\infty}\frac{\sqrt{x^{2}+2x}+3x}{\sqrt{4x^{2}+1}-x+3}=\lim\limits_{x\to -\infty}\frac{\sqrt{x^{2}.\left( 1+\frac{2}{x} \right)}+3x}{\sqrt{x^{2}.\left( 4+\frac{1}{x^{2}} \right)}-x+3}$
$=\lim\limits_{x\to -\infty}\frac{\left| x \right|\sqrt{1+\frac{2}{x}}+3x}{\left| x \right|\sqrt{4+\frac{1}{x^{2}}}-x+3}=\lim\limits_{x\to -\infty}\frac{-x.\sqrt{1+\frac{2}{x}}+3x}{-x.\sqrt{4+\frac{1}{x^{2}}}-x+3}=\lim\limits_{x\to -\infty}\frac{-\sqrt{1+\frac{2}{x}}+3}{-\sqrt{4+\frac{1}{x^{2}}}-1+\frac{3}{x}}=\frac{-2}{3}$