ĐGTD ĐH Bách khoa - Tư duy Toán học - Các dạng vô định

$\begin{array}{l}\underset{x\to 3}{\lim }\frac{\sqrt{x+1}-2}{\sqrt{3x}-3}\\ =\underset{x\to 3}{\lim }\frac{\left(\sqrt{x+1}-2\right)\left(\sqrt{x+1}+2\right)\left(\sqrt{3x}+3\right)}{\left(\sqrt{3x}-3\right)\left(\sqrt{3x}+3\right)\left(\sqrt{x+1}+2\right)}\\ =\underset{x\to 3}{\lim }\frac{(x+1-4)\left(\sqrt{3x}+3\right)}{(3x-9)\left(\sqrt{x+1}+2\right)}\\ =\underset{x\to 3}{\lim }\frac{(x-3)\left(\sqrt{3x}+3\right)}{3(x-3)\left(\sqrt{x+1}+2\right)}\\ =\underset{x\to 3}{\lim }\frac{\sqrt{3x}+3}{3\left(\sqrt{x+1}+2\right)}\\ =\frac{\sqrt{3.3}+3}{3\left(\sqrt{3+1}+2\right)}=\frac{1}{2}\end{array}$

$\underset{x\to 2}{\lim }\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}$

$=\underset{x\to 2}{\lim }\frac{(x-\sqrt{x+2})(x+\sqrt{x+2})(\sqrt{4x+1}+3)}{(\sqrt{4x+1}-3)(\sqrt{4x+1}+3)(x+\sqrt{x+2})}$

$\begin{array}{l}=\underset{x\to 2}{\lim }\frac{(x^2-x-2)(\sqrt{4x+1}+3)}{(4x+1-9)(x+\sqrt{x+2})}\\ =\underset{x\to 2}{\lim }\frac{(x+1)(x-2)(\sqrt{4x+1}+3)}{4(x-2)(x+\sqrt{x+2})}\\ =\underset{x\to 2}{\lim }\frac{(x+1)(\sqrt{4x+1}+3)}{4(x+\sqrt{x+2})}\\ =\frac{(2+1)(\sqrt{4.2+1}+3)}{4(2+\sqrt{2+2})}=\frac{9}{8}\end{array}$

$\begin{array}{l}\underset{x\to 0}{\lim }\frac{1-\sqrt[3]{x+1}}{3x}\\ =\underset{x\to 0}{\lim }\frac{(1-{}^3\sqrt{x+1})(1+{}^3\sqrt{x+1}+{\left({}^3\sqrt{x+1}\right)}^2)}{3x(1+{}^3\sqrt{x+1}+{\left({}^3\sqrt{x+1}\right)}^2)}\\ =\underset{x\to 0}{\lim }\frac{1-(x+1)}{3x(1+{}^3\sqrt{x+1}+{\left({}^3\sqrt{x+1}\right)}^2)}\\ =\underset{x\to 0}{\lim }\frac{-1}{3x(1+{}^3\sqrt{x+1}+{\left({}^3\sqrt{x+1}\right)}^2)}\\ =\underset{x\to 0}{\lim }\frac{-1}{3\left(1+\sqrt[3]{0+1}+{\left(\sqrt[3]{0+1}\right)}^2\right)}=\frac{-1}{9}\\ =\underset{x\to 0}{\lim }\frac{-x}{3x(1+{}^3\sqrt{x+1}+{\left({}^3\sqrt{x+1}\right)}^2)}\end{array}$

$\begin{array}{l}\underset{x\to -\infty }{\lim }(x-1)\sqrt{\frac{x^2}{2x^4+x^2+1}}\\ =\underset{x\to -\infty }{\lim }\left[-\sqrt{\frac{x^2{(x-1)}^2}{2x^4+x^2+1}}\right]\\ =\underset{x\to -\infty }{\lim }\left[-\sqrt{\frac{x^2(x^2-2x+1)}{2x^4+x^2+1}}\right]\\ =\underset{x\to -\infty }{\underset{x\to -\infty }{\lim }\left[-{\sqrt{\frac{x^4-2x^3+x^2}{2x^4+x^2+1}}}^{}\right]}\\ =\underset{x\to -\infty }{\lim }\left[-\sqrt{\frac{1-\frac{2}{x}+\frac{1}{x^2}}{2+\frac{1}{x^2}+\frac{1}{x^4}}}\right]=-\frac{\sqrt{2}}{2}\end{array}$

$\underset{x\to -\infty }{\lim }\left(\sqrt{x^2+1}+x-1\right)$

$=\underset{x\to -\infty }{\lim }\frac{(\sqrt{x^2+1}+x-1)(\sqrt{x^2+1}-x+1)}{\sqrt{x^2+1}-x+1}$

$\begin{array}{l}=\underset{x\to -\infty }{\lim }\frac{x^2+1-{(x-1)}^2}{\sqrt{x^2+1}-x+1}\\ =\underset{x\to -\infty }{\lim }\frac{x^2+1-x^2+2x-1}{\sqrt{x^2+1}-x+1}\\ =\underset{x\to -\infty }{\lim }\frac{2x}{\sqrt{x^2+1}-x+1}\\ =\underset{x\to -\infty }{\lim }\frac{\frac{2x}{x}}{\frac{\sqrt{x^2+1}}{x}-\frac{x}{x}+\frac{1}{x}}\\ =\underset{x\to -\infty }{\lim }\frac{2}{-\sqrt{1+\frac{1}{x^2}}-1+\frac{1}{x}}\\ =\frac{2}{-1-1+0}=-1\end{array}$

$f\left(x\right)=\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}$

Ta có:

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

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

$=\underset{x\to +\infty }{lim}\frac{(x^2+2x+4)-(x^2-2x+4)}{\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}}$

$\begin{array}{l}=\underset{x\to +\infty }{lim}\frac{4x}{\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}}\\ =\underset{x\to +\infty }{lim}\frac{4}{\sqrt{1+\frac{2}{x}+\frac{4}{x^2}}+\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}}=2\end{array}$

$\begin{array}{l}\underset{x\to -\infty }{lim}f\left(x\right)=\underset{x\to -\infty }{lim}\left(\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}\right)\left(\sqrt{x^2+2x+4}-\sqrt{x^2-2x+4}\right)\\ =\underset{x\to -\infty }{lim}\frac{\left(\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}\right)}{\left(\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}\right)}\\ =\underset{x\to -\infty }{lim}\frac{(x^2+2x+4)-(x^2-2x+4)}{\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}}\\ =\underset{x\to -\infty }{lim}\frac{4x}{\sqrt{x^2+2x+4}+\sqrt{x^2-2x+4}}\\ =\underset{x\to -\infty }{lim}\frac{\frac{4x}{x}}{\frac{\sqrt{x^2+2x+4}}{x}+\frac{\sqrt{x^2-2x+4}}{x}}\\ =\underset{x\to -\infty }{lim}\frac{4}{\sqrt{1+\frac{2}{x}+\frac{4}{x^2}}+\sqrt{1-\frac{2}{x}+\frac{4}{x^2}}}=\frac{4}{-1-1}=-2\end{array}$

$\Rightarrow \underset{x\to +\infty }{\lim }f\left(x\right)=-\underset{x\to -\infty }{\lim }f\left(x\right)$

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

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

$\begin{array}{l}\underset{x\to -\infty }{\lim }x\sqrt{\frac{3x+2}{2x^3+x^2-1}}=\underset{x\to -\infty }{\lim }\left(-\sqrt{\frac{x^2\left(3x+2\right)}{2x^3+x^2-1}}\right)=\underset{x\to -\infty }{\lim }\left(-\sqrt{\frac{3x^3+2x^2}{2x^3+x^2-1}}\right)\\ =\underset{x\to -\infty }{\lim }\left(-\sqrt{\frac{3+\frac{2}{x}}{2+\frac{1}{x}-\frac{1}{x^3}}}\right)=-\sqrt{\frac{3}{2}}\end{array}$

Ta có:

$\begin{array}{l}\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1\\ =\sqrt{1+2x}-\sqrt{1+2x}+\sqrt{1+2x}.\sqrt[3]{1+3x}-\sqrt{1+2x}.\sqrt[3]{1+3x}+\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1\\ =\left(\sqrt{1+2x}-1\right)+\sqrt{1+2x}\left(\sqrt[3]{1+3x}-1\right)+\sqrt{1+2x}.\sqrt[3]{1+3x}.\left(\sqrt[4]{1+4x}-1\right)\end{array}$

$\begin{array}{l}\Rightarrow \underset{x\to 0}{\lim }\frac{\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1}{x}\\ =\underset{x\to 0}{\lim }\left(\frac{\sqrt{1+2x}-1}{x}\right)+\underset{x\to 0}{\lim }\left(\sqrt{1+2x}.\frac{\sqrt[3]{1+3x}-1}{x}\right)+\underset{x\to 0}{\lim }\left(\sqrt{1+2x}.\sqrt[3]{1+3x}.\frac{\sqrt[4]{1+4x}-1}{x}\right)\end{array}$

Tính:

$\underset{x\to 0}{\lim }\left(\frac{\sqrt{1+2x}-1}{x}\right)=\underset{x\to 0}{\lim }\frac{\left(\sqrt{1+2x}-1\right)\left(\sqrt{1+2x}+1\right)}{x\left(\sqrt{1+2x}+1\right)}=\underset{x\to 0}{\lim }\frac{2x}{x\left(\sqrt{1+2x}+1\right)}=\underset{x\to 0}{\lim }\frac{2}{\sqrt{1+2x}+1}=\frac{2}{1+1}=1$

$\begin{array}{l}\underset{x\to 0}{lim}\left(\sqrt{1+2x}.\frac{\sqrt[3]{1+3x}-1}{x}\right)\\ =\underset{x\to 0}{lim}\left(\sqrt{1+2x}.\frac{\left(\sqrt[3]{1+3x}-1\right)\left[{\left(\sqrt[3]{1+3x}\right)}^2+\sqrt[3]{1+3x}+1\right]}{x.\left[{\left(\sqrt[3]{1+3x}\right)}^2+\sqrt[3]{1+3x}+1\right]}\right)\\ =\underset{x\to 0}{lim}\left(\sqrt{1+2x}.\frac{3x}{x.\left[{\left(\sqrt[3]{1+3x}\right)}^2+\sqrt[3]{1+3x}+1\right]}\right)\\ =\underset{x\to 0}{lim}\left(\frac{3\sqrt{1+2x}}{\left[{\left(\sqrt[3]{1+3x}\right)}^2+\sqrt[3]{1+3x}+1\right]}\right)=\frac{3.1}{1+1+1}=3\end{array}$

$\underset{x\to 0}{lim}\left(\sqrt{1+2x}.\sqrt[3]{1+3x}.\frac{\sqrt[4]{1+4x}-1}{x}\right)$

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

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

$=\underset{x\to 0}{lim}\frac{4\sqrt{1+2x}.\sqrt[3]{1+3x}}{\left[{\left(\sqrt[4]{1+4x}\right)}^3+{\left(\sqrt[4]{1+4x}\right)}^2+\sqrt[4]{1+4x}+1\right]}=\frac{\mathrm{4.1.1}}{1+1+1+1}=1$

Vậy $\underset{x\to 0}{\lim }\frac{\sqrt{1+2x}.\sqrt[3]{1+3x}.\sqrt[4]{1+4x}-1}{x}=1+1+1=3$