1: \(\lim\limits_{x\rightarrow4}\dfrac{1-x}{\left(x-4\right)^2}=-\infty\) 

vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow4}1-x=1-4=-3< 0\\\lim\limits_{x\rightarrow4}\left(x-4\right)^2=\left(4-4\right)^2=0\end{matrix}\right.\)

2: \(\lim\limits_{x\rightarrow3^+}\dfrac{2x-1}{x-3}=+\infty\)

vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow3^+}2x-1=2\cdot3-1=5>0\\\lim\limits_{x\rightarrow3^+}x-3=3-3>0\end{matrix}\right.\) và x-3>0

3: \(\lim\limits_{x\rightarrow2^+}\dfrac{-2x+1}{x+2}\)

\(=\dfrac{-2\cdot2+1}{2+2}=\dfrac{-3}{4}\)

4: \(\lim\limits_{x\rightarrow1^-}\dfrac{3x-1}{x+1}=\dfrac{3\cdot1-1}{1+1}=\dfrac{2}{2}=1\)

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-\left(x+1\right)}{2x^2-x}=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{x^2+1}-\left(x+1\right)\right)\left(\sqrt{x^2+1}+x+1\right)}{x\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-2x}{x\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-2}{\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)

\(=\dfrac{-2}{\left(0-1\right)\left(\sqrt{1}+1\right)}=1\)

a. \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\lim\limits_{x\rightarrow2}\dfrac{1}{x+2}=\dfrac{1}{4}\)

b. \(\lim\limits_{x\rightarrow3^-}\dfrac{x+3}{x-3}=\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}\)

Do \(\lim\limits_{x\rightarrow3^-}\left(-x-3\right)=-6< 0\)

\(\lim\limits_{x\rightarrow3^-}\left(3-x\right)=0\) và \(3-x>0;\forall x< 3\)

\(\Rightarrow\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}=-\infty\)

a) Áp dụng giới hạn một bên thường dùng, ta có : \(\mathop {\lim }\limits_{x \to {4^ + }} \frac{1}{{x - 4}} =  + \infty \)

b) \(\mathop {\lim }\limits_{x \to {2^ + }} \frac{x}{{2 - x}} = \mathop {\lim }\limits_{x \to {2^+ }} \frac{{ - x}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \left( { - x} \right).\mathop {\lim }\limits_{x \to {2^ + }} \frac{1}{{x - 2}}\)

Ta có: \(\mathop {\lim }\limits_{x \to {2^ + }} \left( { - x} \right) =  - \mathop {\lim }\limits_{x \to {2^ + }} x =  - 2;\mathop {\lim }\limits_{x \to {2^ +}} \frac{1}{{x - 2}} =  +\infty \)

\( \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} \frac{x}{{2 - x}} =  - \infty \)

a) \(\mathop {\lim }\limits_{x \to  - 1} \left( {3{x^2} - x + 2} \right) = \mathop {\lim }\limits_{x \to  - 1} \left( {3{x^2}} \right) - \mathop {\lim }\limits_{x \to  - 1} x + \mathop {\lim }\limits_{x \to  - 1} 2\)

                                                \( = 3\mathop {\lim }\limits_{x \to  - 1} \left( {{x^2}} \right) - \mathop {\lim }\limits_{x \to  - 1} x + \mathop {\lim }\limits_{x \to  - 1} 2 = 3.{\left( { - 1} \right)^2} - \left( { - 1} \right) + 2 = 6\)

b) \(\mathop {\lim }\limits_{x \to 4} \frac{{{x^2} - 16}}{{x - 4}} = \mathop {\lim }\limits_{x \to 4} \frac{{\left( {x - 4} \right)\left( {x + 4} \right)}}{{x - 4}} = \mathop {\lim }\limits_{x \to 4} \left( {x + 4} \right) = \mathop {\lim }\limits_{x \to 4} x + \mathop {\lim }\limits_{x \to 4} 4 = 4 + 4 = 8\)

c) \(\mathop {\lim }\limits_{x \to 2} \frac{{3 - \sqrt {x + 7} }}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {3 - \sqrt {x + 7} } \right)\left( {3 + \sqrt {x + 7} } \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{{3^2} - \left( {x + 7} \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}}\)

                                         \( = \mathop {\lim }\limits_{x \to 2} \frac{{2 - x}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{ - \left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}} = \mathop {\lim }\limits_{x \to 2} \frac{{ - 1}}{{3 + \sqrt {x + 7} }}\)

                                         \( = \frac{{\mathop {\lim }\limits_{x \to 2} \left( { - 1} \right)}}{{\mathop {\lim }\limits_{x \to 2} 3 + \sqrt {\mathop {\lim }\limits_{x \to 2} x + \mathop {\lim }\limits_{x \to 2} 7} }} = \frac{{ - 1}}{{3 + \sqrt {2 + 7} }} =  - \frac{1}{6}\)

a) \(\mathop {\lim }\limits_{x \to  - 3} \left( {4{x^2} - 5x + 6} \right) = 4.{\left( { - 3} \right)^2} - 5.\left( { - 3} \right) + 6 = 57\)

b) \(\mathop {\lim }\limits_{x \to 2} \frac{{2{x^2} - 5x + 2}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 2} \right)\left( {2x - 1} \right)}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \left( {2x - 1} \right) = 2.2 - 1 = 3\)

c) \(\begin{array}{c}\mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x  - 2}}{{{x^2} - 16}} = \mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x  - 2}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = \mathop {\lim }\limits_{x \to 4} \frac{{\sqrt x  - 2}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)\left( {x + 4} \right)}} = \mathop {\lim }\limits_{x \to 4} \frac{1}{{\left( {\sqrt x  + 2} \right)\left( {x + 4} \right)}}\\ = \frac{1}{{\left( {\sqrt 4  + 2} \right)\left( {4 + 4} \right)}} = \frac{1}{{32}}\end{array}\)

Chọn A

\(\lim\limits_{x\rightarrow2}\dfrac{x^2-4}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x+2\right)\left(x-2\right)}{x-2}=\lim\limits_{x\rightarrow2}\left(x+2\right)=4\)

a: \(\lim\limits_{x\rightarrow-2}x^2-7x+4=\left(-2\right)^2-7\cdot\left(-2\right)+4=22\)

b: \(\lim\limits_{x\rightarrow3}\dfrac{x-3}{x^2-9}=\lim\limits_{x\rightarrow3}\dfrac{1}{x+3}=\dfrac{1}{3+3}=\dfrac{1}{6}\)

c: \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-x-8}{3+\sqrt{x+8}}\cdot\dfrac{1}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{-1}{3+\sqrt{x+8}}\)

\(=-\dfrac{1}{6}\)

a) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{9x + 1}}{{3x - 4}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\left( {9 + \frac{1}{x}} \right)}}{{x\left( {3 - \frac{4}{x}} \right)}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{9 + \frac{1}{x}}}{{3 - \frac{4}{x}}} = \frac{{9 + 0}}{{3 - 0}} = 3\)

b) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{7x - 11}}{{2x + 3}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{x\left( {7 - \frac{{11}}{x}} \right)}}{{x\left( {2 + \frac{3}{x}} \right)}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{7 - \frac{{11}}{x}}}{{2 + \frac{3}{x}}} = \frac{{7 - 0}}{{2 + 0}} = \frac{7}{2}\)

c) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {{x^2} + 1} }}{x} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\sqrt {1 + \frac{1}{{{x^2}}}} }}{x} = \mathop {\lim }\limits_{x \to  + \infty } \sqrt {1 + \frac{1}{{{x^2}}}}  = \sqrt {1 + 0}  = 1\)

d) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {{x^2} + 1} }}{x} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x\sqrt {1 + \frac{1}{{{x^2}}}} }}{x} = \mathop {\lim }\limits_{x \to  - \infty }  - \sqrt {1 + \frac{1}{{{x^2}}}}  =  - \sqrt {1 + 0}  =  - 1\)

e) Ta có: \(\left\{ \begin{array}{l}1 > 0\\x - 6 < 0,x \to {6^ - }\end{array} \right.\)

Do đó, \(\mathop {\lim }\limits_{x \to {6^ - }} \frac{1}{{x - 6}} =  - \infty \)                

g) Ta có: \(\left\{ \begin{array}{l}1 > 0\\x + 7 > 0,x \to {7^ + }\end{array} \right.\)

Do đó, \(\mathop {\lim }\limits_{x \to {7^ + }} \frac{1}{{x - 7}} =  + \infty \)

\(\lim\limits_{x\rightarrow a}\dfrac{x^4-a^4}{x^2-a^2}=\lim\limits_{x\rightarrow a}\left(x^2+a^2\right)=2a^2\)