Trắc nghiệm Ôn tập Chương IV-Giới hạn (có đáp án)

Ta có $\left|\frac{\sin \left(n!\right)}{n^2+1}\right|\le \frac{1}{n^2+1}$ mà $\lim \frac{1}{n^2+1}=0$ nên $\lim \frac{\sin \left(n!\right)}{n^2+1}=0$

Chọn đáp án A.

Ta có $0<u_n=\frac{\sqrt{5^n}}{3^n+1}<\frac{{\left(\sqrt{5}\right)}^n}{3^n}={\left(\frac{\sqrt{5}}{3}\right)}^n;\forall n$ 

Lại có $\lim {\left(\frac{\sqrt{5}}{3}\right)}^n=0$ nên theo định lý kẹp có $\lim \frac{\sqrt{5^n}}{3^n+1}=0$

Chọn đáp án C

$$\begin{gathered} \lim {\left(2n+1\right)}^2\left(\frac{3}{n^2+2n}-\frac{1}{n^2+3n-1}\right) \\ =\lim \frac{{\left(2n+1\right)}^2\left(2n^2+7n-3\right)}{\left(n^2+2n\right)\left(n^2+3n-1\right)} \\ =\lim \frac{n^2{\left(2+\frac{1}{n}\right)}^2.n^2\left(2+\frac{7}{n}-\frac{3}{n^2}\right)}{n^2\left(1+\frac{2}{n}\right).n^2\left(1+\frac{3}{n}-\frac{1}{n^2}\right)} \\ =\lim \frac{{\left(2+\frac{1}{n}\right)}^2\left(2+\frac{7}{n}-\frac{3}{n^2}\right)}{\left(1+\frac{2}{n}\right)\left(1+\frac{3}{n}-\frac{1}{n^2}\right)}=\frac{2^2.2}{1.1}=8 \end{gathered}$$

Chọn đáp án D

$$\begin{gathered} \lim \frac{\sqrt{2n+2}-\sqrt{n}}{\sqrt{n}}=\lim \frac{\sqrt{n}.\sqrt{2+\frac{2}{n}}-\sqrt{n}}{\sqrt{n}} \\ =\lim \frac{\sqrt{2+\frac{2}{n}}-1}{1}=\sqrt{2}-1 \end{gathered}$$

Chọn đáp án A

$\lim \frac{\sqrt{2n+1}}{\sqrt{n}+1}=\lim \frac{\sqrt{n}.\sqrt{2+\frac{1}{n}}}{\sqrt{n}.\left(1+\sqrt{\frac{1}{n}}\right)}=\lim \frac{\sqrt{2+\frac{1}{n}}}{1+\sqrt{\frac{1}{n}}}=\sqrt{2}$

Chọn đáp án D

$$\begin{gathered} \lim \frac{4^{n+1}+6^{n+2}}{5^n+8^n}=\lim \frac{{4.4}^n+36.6^n}{5^n+8^n} \\ =\lim \frac{4.{\left(\frac{4}{8}\right)}^n+36.{\left(\frac{6}{8}\right)}^n}{{\left(\frac{5}{8}\right)}^n+1}=\frac{4.0+\text{}36.0}{0+\text{}1}=0 \end{gathered}$$

Chọn đáp án A

$$\begin{gathered} \lim \frac{2n^4-3n^3+2}{n^3+2}=\lim \frac{n^4\left(2-\frac{3}{n}+\frac{2}{n^4}\right)}{n^3\left(1+\frac{2}{n^3}\right)} \\ =\lim \left[n.\frac{2-\frac{3}{n}+\frac{2}{n^4}}{1+\frac{2}{n^3}}\right]=+\infty \end{gathered}$$

Vì $\lim n=+\infty ; \lim \frac{2-\frac{3}{n}+\frac{2}{n^4}}{1+\frac{2}{n^3}}=2$

$$\begin{gathered} \underset{x\to 3^{-}}{\lim }\frac{\left|x-3\right|}{5x-15}=\underset{x\to 3^{-}}{\lim }\frac{-x+3}{5x-15} \\ =\underset{x\to 3^{-}}{\lim }\frac{-(x-3)}{5(x-3)}=\underset{x\to 3^{-}}{\lim }\frac{-1}{5}=\frac{-1}{5} \end{gathered}$$

Chọn đáp án B

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

Chọn đáp án D

Ta có:

$$\begin{gathered} \lim u_n=\lim \left(n-1\right)\sqrt{\frac{2n+2}{n^4+n^2-1}}=\lim \sqrt{\frac{{\left(n-1\right)}^2\left(2n+2\right)}{n^4+n^2-1}} \\ =\lim u_n=\lim \left(n-1\right)\sqrt{\frac{2n+2}{n^4+n^2-1}} \\ =\lim \sqrt{\frac{\frac{2}{n}-\frac{2}{n^2}-\frac{2}{n^3}+\frac{2}{n^4}}{1+\frac{1}{n^2}-\frac{1}{n^4}}}=0. \end{gathered}$$

Chọn đáp án B

$$\begin{gathered} L=\underset{x\to 2^{-}}{\lim }\left(\frac{1}{x-2}-\frac{1}{x^2-4}\right) \\ =\underset{x\to 2^{-}}{\lim }\left(\frac{1}{x-2}-\frac{1}{\left(x-2\right)\left(x+2\right)}\right) \\ =\underset{x\to 2^{-}}{\lim }\frac{x+2-1}{\left(x-2\right)\left(x+2\right)}=\underset{x\to 2^{-}}{\lim }\frac{x+1}{\left(x-2\right)\left(x+2\right)} \end{gathered}$$

Ta có 

$$\begin{gathered} x\to 2^{-}\Rightarrow x<2\Rightarrow x-2<0 \\ \Rightarrow \underset{x\to 2^{-}}{\lim }\left(x-2\right)=0,\underset{x\to 2^{-}}{\lim }\left(x+1\right)=3,\underset{x\to 2^{-}}{\lim }\left(x+2\right)=4 \end{gathered}$$

 Kết luận $L=-\infty$

Chọn đáp án B

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

Ta có $x\to -3^{+}\Rightarrow x>-3\Rightarrow x+3>0\Rightarrow \underset{x\to -3^{+}}{\lim }\left(x+3\right)=0$, và $\underset{x\to -3^{+}}{\lim }\left(2x-1\right)=-7.$

Kết luận $L=-\infty$

Chọn đáp án B.

$$\begin{gathered} \underset{x\to -\infty }{\lim }\frac{2\left|x\right|+3}{\sqrt{x^2+x+5}}=\underset{x\to -\infty }{\lim }\frac{-2x+3}{-x\sqrt{1+\frac{1}{x}+\frac{5}{x^2}}} \\ =\underset{x\to -\infty }{\lim }\frac{-2+\frac{3}{x}}{-\sqrt{1+\frac{1}{x}+\frac{5}{x^2}}}=2 \end{gathered}$$

Chọn đáp án C

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

Chọn đáp án A

Ta có:

$\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{-}}{\lim }\left(x+m\right)=m; \underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\left(x^2+1\right)=1$

Hàm số có giới hạn tại $x=0\iff \underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }f\left(x\right)\iff m=1$

Chọn đáp án D

Đặt 

$$\begin{gathered} u_n=\frac{1}{1.2}+\frac{1}{2.3}+\mathrm{....}+\frac{1}{n\left(n+1\right)} \\ =1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\mathrm{...}+\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1} \end{gathered}$$

$\Rightarrow \lim u_n=\lim \left(1-\frac{1}{n+1}\right)=1$

Chọn đáp án B

Tại điểm x= -1 ta có:

$$\begin{gathered} \underset{x\to {\left(-1\right)}^{-}}{\lim }f\left(x\right)=\underset{x\to {\left(-1\right)}^{-}}{\lim }\left(3x+b\right)=-3+\text{}b=f(-1) \\ \underset{x\to {\left(-1\right)}^{+}}{\lim }f\left(x\right)=\underset{x\to {\left(-1\right)}^{+}}{\lim }\left(x+a\right)=-1+a \end{gathered}$$

Hàm số có giới hạn tại x= -1 khi và chỉ khi $\underset{x\to {\left(-1\right)}^{-}}{\lim }f\left(x\right)=\underset{x\to {\left(-1\right)}^{+}}{\lim }f\left(x\right)$

Điều này tương đương với $-3+b=-1+a\Rightarrow a-b=-2$

Chọn đáp án B.

Ta có:

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

Hàm số có giới hạn tại x= -1 khi và chỉ khi $\underset{x\to -1^{-}}{\lim }h\left(x\right)=\underset{x\to -1^{+}}{\lim }h\left(x\right)$

$3=m+1+\text{\hspace{0.17em}}{m}^2\iff m^2+m-2=0\iff \left[\begin{array}{l}m=1\\ m=-2\end{array}\right.$

Chọn đáp án C

Ta có :

$\left\{\begin{array}{l}\underset{x\to 3^{-}}{\lim }f\left(x\right)=\underset{x\to 3^{-}}{\lim }m=m\\ \underset{x\to 3^{+}}{\lim }f\left(x\right)=\underset{x\to 3^{+}}{\lim }\frac{3-x}{\sqrt{x+1}-2}=\underset{x\to 3^{+}}{\lim }\frac{(3-x).\text{}(\sqrt{x+1}+2)}{x-3}=\underset{x\to 3^{+}}{\lim }-(\sqrt{x+1}+2)=-4\end{array}\right.$

Vậy để tồn tại $\underset{x\to 3}{\lim }f\left(x\right)$ khi và chỉ khi :

$\underset{x\to 3^{-}}{\lim }f\left(x\right)=\underset{x\to 3^{+}}{\lim }f\left(x\right)\iff m=-4$ 

Chọn đáp án C

Với x= 1 ta có :$f\left(1\right)=k^2$

Với $x\ne 1$ ta có :

$\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{-}}{\lim }\left(x^2+3\right)=4; \underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }{\left(x+1\right)}^2=4$ suy ra $\underset{x\to 1}{\lim }f\left(x\right)=4$

Vậy để hàm số gián đoạn tại x= 1 khi $\underset{x\to 1}{\lim }f\left(x\right)\ne k^2\iff k^2\ne 4\iff k\ne \pm 2$

Chọn đáp án A

Ta có : $f\left(-1\right)=1$ và $\underset{x\to {\left(-1\right)}^{-}}{\lim }f\left(x\right)=\underset{x\to {\left(-1\right)}^{-}}{\lim }(2x+\text{}3)=2.(-1)\text{}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3=1$

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

Suy ra : $\underset{x\to {\left(-1\right)}^{+}}{\lim }f\left(x\right)\ne \underset{x\to {\left(-1\right)}^{-}}{\lim }f\left(x\right)$

Vậy hàm số gián đoạn tại x= -1.

Chọn đáp án C

$\underset{x\to 2}{\lim }\sqrt{\frac{4x^2-3x}{\left(2x-1\right)\left(x^3-2\right)}}=\sqrt{\frac{{4.2}^2-3.2}{\left(2.2-1\right)\left(2^3-2\right)}}=\sqrt{\frac{10}{18}}=\frac{\sqrt{5}}{3}.$

Chọn đáp án D

Ta có $4x^5-3x^3+x+1=x^5\left(4-\frac{3}{x^2}+\frac{1}{x^4}+\frac{1}{x^5}\right)$

Vì $\underset{x\to -\infty }{\lim }{x}^5=-\infty ;\underset{x\to -\infty }{\lim }\left(4-\frac{3}{x^2}+\frac{1}{x^4}+\frac{1}{x^5}\right)=4>0$

Vậy: $\underset{x\to -\infty }{\lim }\left(4x^5-3x^3+x+1\right)=-\infty$

Chọn đáp án D

$A=\underset{x\to 2}{\lim }\frac{3x+m}{x+2}=\frac{6+m}{4},$

Mà $A=5\Rightarrow \frac{6+m}{4}=5\iff 6+\text{}m=20\iff m=14$

Chọn đáp án B

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

Vì: $\underset{x\to -1}{\lim }{x}^2=1>0;\underset{x\to -1}{\lim }{\left(x+1\right)}^2=0$ và ${\left(x+1\right)}^2>0,\forall x\ne -1$

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

Chọn đáp án D