34 câu Trắc nghiệm Giới hạn của hàm số có đáp án (phần 2)

Chọn B

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

Vì: 

$\begin{array}{l}\underset{x\to -\infty }{\lim }{x}^2=+\text{}\infty ;\\ \underset{x\to -\infty }{\lim }\left(-\sqrt{4+\frac{1}{x^2}}-1\right)= -\sqrt{4 +0}-1 =-3<0\end{array}$

Chọn B

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

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

Chọn C

$\begin{array}{l}C=\underset{x\to 3}{\lim }\frac{\sqrt{2x+3}-x}{x^2-4x+3}=\underset{x\to 3}{\lim }\frac{2x +3-x^2}{(x- 3). (x- 1). (\sqrt{2x+3}+x) } \\ =\underset{x\to 3}{\lim }\frac{-(x-3)(x+1)}{(x-3)(x-1)\left(\sqrt{2x+3}+x\right)}\\ =\underset{x\to 3}{\lim }\frac{-(x+1)}{(x-1)\left(\sqrt{2x+3}+x\right)}\\ =\frac{-(3+\text{}1)}{(3-1).(\sqrt{2.3+\text{}3}+\text{}3)}=\frac{-1}{3}\end{array}$

Chọn C.

Ta có:

$\begin{array}{l}\underset{x\to 2^{+}}{\lim }f\left(x\right)\\ =\underset{x\to 2^{+}}{\lim }(x^2+ax+2)=2a+6\end{array}$.

$\underset{x\to 2^{-}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }(2x^2-x+1)=7$.

Hàm số có giới hạn khi$x\to 2\iff \underset{x\to 2^{+}}{\lim }f\left(x\right)=\underset{x\to 2^{-}}{\lim }f\left(x\right)$$\iff 2a+6=7\iff a=\frac{1}{2}$.

Vậy $a=\frac{1}{2}$ là giá trị cần tìm.

Chọn C

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

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

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

Vậy $A=2-\frac{2}{3}=\frac{4}{3}$

Chọn  B

$\underset{x\to 0^{-}}{\lim }\left(\frac{1}{x^2}-\frac{2}{x^3}\right)=\underset{x\to 0^{-}}{\lim }\left(\frac{x-2}{x^3}\right)$

$\underset{x\to 0^{-}}{\lim }\left(x-2\right)=-2<0;\underset{x\to 0^{-}}{\lim }{x}^3 = 0 \hspace{0.17em}$

khi $x\to 0^{-}\Rightarrow x<0\Rightarrow x^3<0$

vậy $\underset{x\to 0^{-}}{\lim }\left(\frac{x-2}{x^3}\right)=+\infty$

Chọn C.

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

Chọn A

$\begin{array}{l}\left.\begin{array}{l}\underset{x\to 3^{+}}{\lim }\frac{\left|x-3\right|}{x-3}=\underset{x\to 3^{+}}{\lim }\frac{x-3}{x-3}=1\\ \underset{x\to 3^{-}}{\lim }\frac{\left|x-3\right|}{x-3}=\underset{x\to 3^{-}}{\lim }\frac{-x+3}{x-3}=-1\end{array}\right\}\\ \Rightarrow \underset{x\to 3^{+}}{\lim }\frac{\left|x-3\right|}{x-3}\ne \underset{x\to 3^{-}}{\lim }\frac{\left|x-3\right|}{x-3}\end{array}$

Vậy không tồn tại giới hạn trên

Chọn A

$\begin{array}{l}\frac{1}{x^3-1}-\frac{1}{x-1}\\ =\frac{1}{(x-1).(x^2+x+1)}-\text{}\frac{1}{x-1}\\ =\frac{1-(x^2+\text{}x+1)}{(x-1).(x^2+x+1)}\\ =\frac{-x^2-x}{(x-1).(x^2+x+1)}\\ =\frac{-x^2-x}{x^3-1}\end{array}$

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

$\underset{x\to 1^{+}}{\lim }\left(-x^2-x\right)=-2; \underset{x\to 1^{+}}{\lim }\left(x^3-1\right)= 0$

khi $x\to 1^{+}\Rightarrow x>1\Rightarrow x^3-1>0$

vậy $\underset{x\to 1^{+}}{\lim }f\left(x\right)=-\infty$

Chọn A

ta có:

Mà và 

Vậy 

chọn đáp án C

Ta có

và 

với mọi $x\not\equiv 4$ nên

Chọn D

Chọn B

Chọn B

chọn đáp án A

Với mọi x> 2 ta có 

Do đó

chọn B

Ta có 

vì 

với mọi x> 2 nên 

Chọn B

Chọn  A

Chọn C

$$\begin{gathered} \underset{x\to 0^{+}}{\lim }\frac{2x+7\sqrt{x}}{5x-\sqrt{x}}=\underset{x\to 0^{+}}{\lim }\frac{\sqrt{x}\left(2\sqrt{x}+7\right)}{\sqrt{x}\left(5\sqrt{x}-1\right)} \\ =\underset{x\to 0^{+}}{\lim }\frac{\left(2\sqrt{x}+7\right)}{\left(5\sqrt{x}-1\right)}=\frac{2.0+7}{5.0- 1} =-7 \end{gathered}$$

Chọn B

Chọn D

$\underset{x\to 5}{\lim }\frac{x^2-12x+35}{3x-15}=\underset{x\to 5}{\lim }\frac{\left(x-5\right)\left(x-7\right)}{3\left(x-5\right)}=\underset{x\to 5}{\lim }\frac{x-7}{3}=-\frac{2}{3}$

Chọn C

Chọn C

Chọn B

$$\begin{gathered} \underset{x\to +\infty }{\lim }x\left(\sqrt{x^2+16}-x\right)=\underset{x\to +\infty }{\lim }x\frac{x^2+16-x^2}{\sqrt{x^2+16}+x} \\ =\underset{x\to +\infty }{\lim }\frac{16x}{\sqrt{x^2+16}+x}=\underset{x\to +\infty }{\lim }\frac{16}{\sqrt{1+{\displaystyle \frac{16}{x^2}}}+1} \\ =\frac{16}{1+\hspace{0.17em}1}=8 \end{gathered}$$

Chọn C

$$\begin{gathered} \underset{t\to a}{\lim }\frac{t^4-a^4}{t-a} = \underset{t\to a}{\lim }\frac{(t^2-a^2).( t^2 +\hspace{0.17em}{a}^2)}{t-a} = \underset{t\to a}{\lim }\frac{(t- a).(t+ a) .( t^2 +\hspace{0.17em}{a}^2)}{t-a} \\ =\underset{t\to a}{\lim }\frac{\left(t-a\right)\left(t^3+a{t}^2+a^{2}t+a^3\right)}{t-a}= a^3 +\hspace{0.17em}a.a^2+\hspace{0.17em}{a}^2.a + a^3=4a^3 \end{gathered}$$