10 câu Trắc nghiệm Giới hạn của hàm số có đáp án (Vận dụng)

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

Ta có

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

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

Đáp án cần chọn là: D

$\begin{array}{l}\underset{x\to +\infty }{\lim }\left(\sqrt{x^2+x}-\sqrt[3]{x^3-x^2}\right)\\ =\underset{x\to +\infty }{\lim }\left(\sqrt{x^2+x}-x+x-\sqrt[3]{x^3-x^2}\right)\\ =\underset{x\to +\infty }{\lim }\left(\frac{x}{\sqrt{x^2+x}+x}+\frac{x^2}{x^2+x\sqrt[3]{x^3-x^2}+\sqrt[3]{{\left(x^3-x^2\right)}^2}}\right)\\ =\underset{x\to +\infty }{\lim }\left(\frac{1}{\sqrt{1+\text{}\frac{1}{x}}+1\text{\hspace{0.17em}\hspace{0.17em}}}+\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{1+\text{}\sqrt[3]{1-\frac{1}{x}}+\text{}\sqrt[3]{{(1-\frac{1}{x})}^2}\text{\hspace{0.17em}\hspace{0.17em}}}\right)\\ =\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\end{array}$

Đáp án cần chọn là: A

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

Đáp án cần chọn là: A

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

Tính 

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

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

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

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

Đáp án cần chọn là: D

Vì $\underset{x\to -\infty }{\lim }x=-\infty$ nên $\underset{x\to -\infty }{\lim }\left(\sqrt{2x^2+1}+ax\right)=\underset{x\to -\infty }{\lim }x\left(-\sqrt{2+\frac{1}{x^2}}+a\right)=+\infty$

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

Đáp án cần chọn là: B

Đặt $x=\frac{1}{y}$ khi $x\to +\infty :y\to 0$

$\begin{array}{l}\underset{x\to +\infty }{\lim }\left(\sqrt[n]{\left(x+1\right)\left(x+2\right)\mathrm{...}\left(x+n\right)}-x\right)\\ =\underset{x\to 0}{\lim }\left(\sqrt[n]{\left(\frac{1}{y}+1\right)\left(\frac{1}{y}+2\right)\mathrm{...}\left(\frac{1}{y}+n\right)}-\frac{1}{y}\right)\\ =\underset{x\to 0}{\lim }\frac{\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+ny)}-1}{y}\end{array}$

$\begin{array}{l}*\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+ny)}-1\\ =\sqrt[n]{1+y}-\sqrt[n]{1+y}+\sqrt[n]{(1+y)(1+2y)}-\sqrt[n]{(1+y)(1+2y)+\mathrm{...}}\\ -\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+(n-1\left)y\right)}+\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+ny)}-1\\ =\left(\sqrt[n]{1+y}-1\right)+\sqrt[n]{1+y}\left(\sqrt[n]{1+2y}-1\right)+\mathrm{...}+\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+(n-1\left)y\right)}\left(\sqrt[n]{1+ny}-1\right)\\ \Rightarrow \underset{y\to 0}{\lim }\frac{\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+ny)}-1}{y}\\ =\underset{y\to 0}{\lim }\left[\frac{\left(\sqrt[n]{1+y}-1\right)}{y}\right]+\underset{y\to 0}{\lim }\left[\sqrt[n]{1+y}\frac{\left(\sqrt[n]{1+2y}-1\right)}{y}\right]+\mathrm{...}\\ +\underset{y\to 0}{\lim }\left[\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+(n-1\left)y\right)}.\frac{\left(\sqrt[n]{1+ny}-1\right)}{y}\right]\end{array}$

Tổng quát

$\begin{array}{l}\underset{y\to 0}{\lim }\left[\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+(k-1)y}.\frac{\sqrt[n]{1+ky}-1}{y}\right]\\ =\underset{y\to 0}{\lim }\left[\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+(k-1)y}.\frac{\left(\sqrt[n]{1+ky}-1\right)\left[{\left(\sqrt[n]{1+ky}\right)}^{n-1}+{\left(\sqrt[n]{1+ky}\right)}^{n-2}+\mathrm{...}+1\right]}{y\left[{\left(\sqrt[n]{1+ky}\right)}^{n-1}+{\left(\sqrt[n]{1+ky}\right)}^{n-2}+\mathrm{...}+1\right]}\right]\\ =\underset{y\to 0}{\lim }\frac{(1+ky-1).\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+(k-1\left)y\right)}}{{\left(\sqrt[n]{1+ky}\right)}^{n-1}+{\left(\sqrt[n]{1+ky}\right)}^{n-2}+\mathrm{...}+1}=\frac{k}{n}\end{array}$

Khi đó

$\begin{array}{l}\underset{y\to 0}{\lim }\frac{\sqrt[n]{(1+y)(1+2y)\mathrm{...}(1+ny)}-1}{y}=\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+\mathrm{...}+\frac{n}{n}\\ =\frac{1+2+3+\mathrm{...}+n}{n}=\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}\end{array}$

Đáp án cần chọn là: B

$\begin{array}{l}\underset{x\to -\sqrt{3}}{\lim }\frac{2(x^3+3\sqrt{3})}{3-x^2}\\ =\underset{x\to -\sqrt{3}}{\lim }\frac{2(x+\sqrt{3})(x^2-\sqrt{3}x+3)}{\left(\sqrt{3}-x\right)\left(\sqrt{3}+x\right)}\\ =\underset{x\to -\sqrt{3}}{\lim }\frac{2(x^2-\sqrt{3}x+3)}{\sqrt{3}-x}\\ =\frac{2\left[{\left(-\sqrt{3}\right)}^2-\sqrt{3}.(-\sqrt{3})+3\right]}{\sqrt{3}-(-\sqrt{3})}=\frac{18}{2\sqrt{3}}=3\sqrt{3}\\ \Rightarrow \left\{\begin{array}{l}a=3\\ b=0\end{array}\right.\Rightarrow a^2+b^2=9\end{array}$

Đáp án cần chọn là: A

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

Đáp án cần chọn là: A

Ta có:

$\begin{array}{l}\underset{x\to 0}{\lim }\frac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}=\underset{x\to 0}{\lim }\left(\frac{2\sqrt{1+x}-2}{x}+\frac{2-\sqrt[3]{8-x}}{x}\right)\\ =\underset{x\to 0}{\lim }\left(\frac{(2\sqrt{1+x}-2).(2\sqrt{1+x}+2)}{x.(2\sqrt{1+x}+2)}+\frac{(2-\sqrt[3]{8-x}).(4+2\sqrt[3]{8-x}+\sqrt[3]{{\left(8-x\right)}^2})\text{}}{x.(4+2\sqrt[3]{8-x}+\sqrt[3]{{\left(8-x\right)}^2})}\right)\\ =\underset{x\to 0}{\lim }\left(\frac{4(1+\text{}x)-4}{x(2\sqrt{x+1}+2)}+\frac{8-(8-x)}{x.\text{[}4+2\sqrt[3]{8-x}+\sqrt[3]{{(8-x)}^2}\text{]}}\right)\\ =\underset{x\to 0}{\lim }\frac{4}{2\sqrt{x+1}+2}+\frac{1}{4+2\sqrt[3]{8-x}+\sqrt[3]{{(8-x)}^2}}\\ =\frac{4}{2.1+2}+\frac{1}{4+\text{}4+\text{}4}=\frac{13}{12}\end{array}$

Đáp án cần chọn là: B