30 câu Dạng 3: Tìm giới hạn của hàm số dạng vô định có đáp án

Hướng dẫn giải

$A=\underset{x\to -\infty }{\lim }\frac{\sqrt[3]{3x^3+1}-\sqrt{2x^2+x+1}}{\sqrt[4]{4x^4+2}}=\underset{x\to -\infty }{\lim }\frac{x\sqrt[3]{3+\frac{1}{x^3}}+x\sqrt{2+\frac{1}{x}+\frac{1}{x^2}}}{x\sqrt[4]{4+\frac{2}{x^4}}}=-\frac{\sqrt[3]{3}+\sqrt{2}}{\sqrt{2}}$

Hướng dẫn giải

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

Ta có: $B=\underset{x\to -\infty }{\lim }\frac{\sqrt{4x^2+x}+\sqrt[6]{64x^6+x-1}}{\sqrt[4]{x^4+3}}=\underset{x\to -\infty }{\lim }\frac{-\sqrt{4+\frac{1}{x}}-\sqrt[6]{64+\frac{1}{x^5}-\frac{1}{x^6}}}{-\sqrt[4]{1+\frac{3}{x^4}}}=4$

Ta có: $\underset{x\to +\infty }{\lim }\frac{x^{14}+7}{x^{14}-1}=\underset{x\to +\infty }{\lim }\frac{1+\frac{7}{x^{14}}}{1-\frac{1}{x^{14}}}=1$

Ta có: $C=\underset{x\to -\infty }{\lim }\frac{2x-\sqrt{9x^2+2}}{5x+\sqrt{x^2+1}}=\underset{x\to -\infty }{\lim }\frac{2+\sqrt{9+\frac{2}{x^2}}}{5-\sqrt{1+\frac{1}{x^2}}}=\frac{5}{4}$

Ta có: $\underset{x\to +\infty }{\lim }f\left(x\right)=\underset{x\to +\infty }{\lim }\sqrt{\frac{\frac{1}{x^{2017}}+\frac{2020}{x^{2019}}}{2+\frac{1}{x^{2017}}}}=0$

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

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

Ta có: $\underset{x\to -\infty }{\lim }\frac{\sqrt{x^4-x^2+3}}{2\left|x\right|\sqrt{4x^2+5}}=\underset{x\to \infty }{\lim }\frac{x^2\sqrt{1-\frac{1}{x^2}+\frac{3}{x^4}}}{2x^2\sqrt{4+\frac{5}{x^2}}}=\underset{x\to \infty }{\lim }\frac{\sqrt{1-\frac{1}{x^2}+\frac{3}{x^4}}}{2\sqrt{4+\frac{5}{x^2}}}=\frac{1}{4}$

Ta có: $\underset{x\to +\infty }{\lim }\frac{x^4+8x\sqrt{x+2}+1}{x^3+2x^2+x+2}=\underset{x\to +\infty }{\lim }\frac{x^4\left(1+\frac{8}{x^2\sqrt{x}}\sqrt{1+\frac{2}{x}}+\frac{1}{x^4}\right)}{x^3\left(1+\frac{2}{x}+\frac{1}{x^2}+\frac{2}{x^3}\right)}=\underset{x\to +\infty }{\lim }\frac{x\left(1+\frac{8}{x^2\sqrt{x}}\sqrt{1+\frac{2}{x}}+\frac{1}{x^4}\right)}{\left(1+\frac{2}{x}+\frac{1}{x^2}+\frac{2}{x^3}\right)}=+\infty$

Vì $\underset{x\to +\infty }{\lim }\frac{\left(1+\frac{8}{x^2\sqrt{x}}\sqrt{1+\frac{2}{x}}+\frac{1}{x^4}\right)}{\left(1+\frac{2}{x}+\frac{1}{x^2}+\frac{2}{x^3}\right)}=1>0$ và $\underset{x\to +\infty }{\lim }x=+\infty$

Ta có: $E=\underset{x\to +\infty }{\lim }\frac{\sqrt{x^2-x+1}-2x}{x+1}=\underset{x\to +\infty }{\lim }\frac{\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}-2}{1+\frac{1}{x}}=-1$

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

Vì $\underset{x\to -\infty }{\lim }\frac{-\sqrt{4+\frac{1}{x^2}}+1}{\sqrt[3]{4+\frac{1}{x^3}}+2}=\frac{-1}{\sqrt[3]{4}+2}<0$ và $\underset{x\to -\infty }{\lim }x=-\infty$

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

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

Ta có: $N=\underset{x\to +\infty }{\lim }\frac{x^2\left(2+\frac{3}{x^2}\right)}{x^2\left(\sqrt[3]{{\left(8+\frac{2}{x^2}\right)}^2}+2\sqrt[3]{8+\frac{2}{x^2}}+4+\frac{2}{x^2}\right)}=\underset{x\to +\infty }{\lim }\frac{2+\frac{3}{x^2}}{\sqrt[3]{{\left(8+\frac{2}{x^2}\right)}^2}+2\sqrt[3]{8+\frac{2}{x^2}}+4+\frac{2}{x^2}}=\frac{1}{6}$

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

$=\underset{x\to -\infty }{\lim }-\frac{\sqrt[4]{16+\frac{3}{x^3}+\frac{1}{x^4}}+\sqrt{4+\frac{2}{x^2}}}{3+\frac{1}{x}}=-\frac{4}{3}$

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

$=\underset{x\to -\infty }{\lim }-\frac{\sqrt[3]{3+\frac{1}{x^3}}+\sqrt{2+\frac{1}{x}+\frac{1}{x^2}}}{\sqrt[4]{4+\frac{2}{x^3}}}=-\frac{\sqrt[3]{3}+\sqrt{2}}{\sqrt{2}}$

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

Ta có: $A=\underset{x\to +\infty }{\lim }\frac{x^{2023}{\left(2+\frac{1}{x}\right)}^3{\left(1+\frac{2}{x}\right)}^{2020}}{x^{2023}\left(\frac{3}{x^4}-2\right){\left(\frac{1}{x}-1\right)}^{2019}}=\underset{x\to +\infty }{\lim }\frac{{\left(2+\frac{1}{x}\right)}^3{\left(1+\frac{2}{x}\right)}^{2020}}{\left(\frac{3}{x^4}-2\right){\left(\frac{1}{x}-1\right)}^{2019}}=4$

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