24 câu Dạng 4: Tìm số hạng của hàm số dạng vô định có đáp án

Hướng dẫn giải

$B=\underset{x\to -\infty }{\lim }\frac{\sqrt{4x^2-3x+4}+2x}{\sqrt{x^2+x+1}+x}=\underset{x\to -\infty }{\lim }\frac{\left(4-3x\right)\left(\sqrt{x^2+x+1}-x\right)}{\left(x+1\right)\left(\sqrt{4x^2-3x+4}-2x\right)}=-\frac{3}{2}$

Hướng dẫn giải

Ta có $A=\underset{x\to +\infty }{\lim }\left(\left(\sqrt[3]{x^3+2x^2+1}-x\right)-2\left(\sqrt{x^2-x}-x\right)\right)$

$=\underset{x\to +\infty }{\lim }\left(\frac{2x^2+1}{\sqrt[3]{{\left(x^3+2x^2+1\right)}^2}+x\sqrt[3]{x^3+2x^2+1}+x^2}+\frac{2x}{\sqrt{x^2-x}+x}\right)=\frac{5}{3}$

Ta có: $A=\underset{x\to +\infty }{\lim }\left(\sqrt{x^2-3x+1}-x\right)\iff A=\underset{x\to +\infty }{\lim }\frac{-3x+1}{\sqrt{x^2-3x+1}+x}\iff A=-\frac{3}{2}$

Vậy $A=\underset{x\to +\infty }{\lim }\left(\sqrt{x^2-3x+1}-x\right)=\frac{-3}{2}$

Ta có : $B=\underset{x\to -\infty }{\lim }\left(2x+\sqrt{4x^2-x+1}\right)\iff B=\underset{x\to -\infty }{\lim }\frac{-x+1}{\sqrt{4x^2-x+1}-2x}\iff B=\frac{1}{4}$

Vậy $B=\underset{x\to -\infty }{\lim }\left(2x+\sqrt{4x^2-x+1}\right)=\frac{1}{4}$

Ta có:$D=\underset{x\to +\infty }{\lim }x\left(\sqrt{9x^2+1}-3x\right)=\underset{x\to +\infty }{\lim }\frac{x}{\sqrt{9x^2+1}+3}=\frac{1}{6}$

Ta có $G=\underset{x\to -\infty }{\lim }\left(\sqrt[3]{x^3-3x^2}+\sqrt{x^2-2x}\right)\iff G=\underset{x\to -\infty }{\lim }\left(\sqrt[3]{x^3-3x^2}-x\right)+\underset{x\to -\infty }{\lim }\left(x+\sqrt{x^2-2x}\right)$

$\iff G=\underset{x\to -\infty }{\lim }\frac{-3x^2}{\sqrt[3]{{\left(x^3-3x^2\right)}^2}+x\sqrt[3]{x^3-3x}+x^2}+\underset{x\to -\infty }{\lim }\frac{2x}{x-\sqrt{x^2-2x}}\iff G=-1+1=0$

Ta có: $H=\underset{x\to +\infty }{\lim }\left(\sqrt[4]{16x^4+3x+1}-\sqrt{4x^2+2}\right)$

$\iff H=\underset{x\to +\infty }{\lim }\left(\sqrt[4]{16x^4+3x+1}-2x\right)+\underset{x\to +\infty }{\lim }\left(2x-\sqrt{4x^2+2}\right)$

$=\underset{x\to +\infty }{\lim }\frac{3x+1}{\sqrt[4]{{\left(16x^4+3x+1\right)}^3}+2x.\sqrt[4]{{\left(16x^4+3x+1\right)}^2}+4x^2\sqrt[4]{16x^4+3x+1}+8x^3}+\underset{x\to +\infty }{\lim }\frac{-2}{2x+\sqrt{4x^2+2}}\iff H=0$

Ta có: $I=\underset{x\to +\infty }{\lim }\frac{\sqrt[3]{x^6+x+1}-4\sqrt{x^4+2x-1}}{{\left(2x+3\right)}^2}=\underset{x\to +\infty }{\lim }\frac{\sqrt[3]{1+\frac{1}{x^5}+\frac{1}{x^6}}-4\sqrt{1+\frac{2}{x^3}-\frac{1}{x^4}}}{{\left(2+\frac{3}{x}\right)}^2}=-\frac{3}{4}$

Suy ra $a+b=7$

Ta có: $J=\underset{x\to +\infty }{\lim }\left(\sqrt{x^2+x+1}-2\sqrt[3]{x^3+x^2-1}+x\right)=\underset{x\to +\infty }{\lim }\left(\sqrt{x^2+x+1}-x\right)+\underset{x\to +\infty }{\lim }2\left(x-\sqrt[3]{x^3+x^2-1}\right)$

$=\underset{x\to +\infty }{\lim }\frac{x+1}{\sqrt{x^2+x+1}+x}+\underset{x\to +\infty }{\lim }2\frac{-x^2+1}{x^2+x\sqrt[3]{x^3+x^2-1}+\sqrt[3]{{\left(x^3+x^2-1\right)}^2}}=\frac{1}{2}-\frac{2}{3}=-\frac{1}{6}$

suy ra a+b =7

Ta có: $K=\underset{x\to +\infty }{\lim }x\left(\sqrt{x^2+2x}-\sqrt[3]{x^3+3x^2}\right)=\underset{x\to +\infty }{\lim }x\left(\sqrt{x^2+2x}-x-1\right)+\underset{x\to +\infty }{\lim }x\left(x+1-\sqrt[3]{x^3+3x^2}\right)$

$=\underset{x\to +\infty }{\lim }\frac{-x}{\sqrt{x^2+2x}+\left(x+1\right)}+\underset{x\to +\infty }{\lim }\frac{3x^2+x}{{\left(x+1\right)}^2+\left(x+1\right)\sqrt[3]{x^3+3x}+\sqrt[3]{{\left(x^3+3x\right)}^2}}=-\frac{1}{2}+1=\frac{1}{2}$

Suy ra $a+b=3$

Ta có: $L=\underset{x\to -\infty }{\lim }\left(\sqrt{4x^2+ax+12}+2x\right)=\underset{x\to -\infty }{\lim }\frac{ax+12}{\sqrt{4x^2+ax+12}-2x}=-\frac{a}{4}$

Suy ra $-\frac{a}{4}=5\iff a=-20$

Ta có: $\underset{x\to -\infty }{\lim }\left(\sqrt{4x^2-ax}+\sqrt[3]{8x^3+b{x}^2+5}\right)=\underset{x\to -\infty }{\lim }\left(\sqrt{4x^2-ax}+2x\right)+\underset{x\to -\infty }{\lim }\left(\sqrt[3]{8x^3+b{x}^2+5}-2x\right)$

$=\underset{x\to -\infty }{\lim }\frac{-ax}{\sqrt{4x^2-ax}-2x}+\underset{x\to -\infty }{\lim }\frac{b{x}^2+5}{\sqrt[3]{{\left(8x^3+b{x}^2+5\right)}^2}+2x\sqrt[3]{8x^3+b{x}^2+5}+4x^2}=\frac{a}{4}+\frac{b}{12}$

Ta có $\frac{2}{3}=\frac{a}{4}+\frac{b}{12}\ge 2\sqrt{\frac{a}{4}.\frac{b}{12}}\iff ab\le \frac{16}{3}$

Ta có $\underset{x\to -\infty }{\lim }\left(\sqrt{2x^2-3x+1}+x\sqrt{2}\right)=\underset{x\to -\infty }{\lim }\frac{-3x+1}{\sqrt{2x^2-3x+1}-x\sqrt{2}}=\frac{3}{4}\sqrt{2}$

Suy ra a+ b =7

$\underset{x\to 0}{\lim }\frac{\sqrt[3]{ax+1}-\sqrt{1-bx}}{x}=\underset{x\to 0}{\lim }\left[\frac{\sqrt[3]{ax+1}-1}{x}+\frac{1-\sqrt{1-bx}}{x}\right]$

$=\underset{x\to 0}{\lim }\left[\frac{ax}{x\left(\sqrt[3]{{\left(ax+1\right)}^2}+\sqrt[3]{ax+1}+1\right)}+\frac{bx}{x\left(1+\sqrt{1-bx}\right)}\right]$

$=\underset{x\to 0}{\lim }\left[\frac{a}{\sqrt[3]{{\left(ax+1\right)}^2}+\sqrt[3]{ax+1}+1}+\frac{b}{1+\sqrt{1-bx}}\right]=\frac{a}{3}+\frac{b}{2}$

Ta có $\underset{x\to +\infty }{\lim }\left(\sqrt{a{x}^2+bx}-cx\right)=\underset{x\to +\infty }{\lim }\frac{{\left(a-c\right)}^2{x}^2+bx}{\sqrt{a{x}^2+bx}+cx}$