a: \(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{2x+10}-4}{3x-9}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{2x+10-16}{3x-9}\cdot\dfrac{1}{\sqrt{2x+10}+4}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{2\left(x-3\right)}{3\left(x-3\right)\cdot\left(\sqrt{2x+10}+4\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{2}{3\left(\sqrt{2x+10}+4\right)}\)

\(=\dfrac{2}{3\cdot\sqrt{6+10}+3\cdot4}=\dfrac{2}{3\cdot4+3\cdot4}=\dfrac{2}{24}=\dfrac{1}{12}\)

b: \(\lim\limits_{x\rightarrow7}\dfrac{\sqrt{4x+8}-6}{x^2-9x+14}\)

\(=\lim\limits_{x\rightarrow7}\dfrac{4x+8-36}{\sqrt{4x+8}+6}\cdot\dfrac{1}{\left(x-2\right)\left(x-7\right)}\)

\(=\lim\limits_{x\rightarrow7}\dfrac{4x-28}{\left(\sqrt{4x+8}+6\right)\cdot\left(x-2\right)\left(x-7\right)}\)

\(=\lim\limits_{x\rightarrow7}\dfrac{4}{\left(\sqrt{4x+8}+6\right)\left(x-2\right)}\)

\(=\dfrac{4}{\left(\sqrt{4\cdot7+8}+6\right)\left(7-2\right)}\)

\(=\dfrac{4}{5\cdot12}=\dfrac{4}{60}=\dfrac{1}{15}\)

c: \(\lim\limits_{x\rightarrow5}\dfrac{x^2-8x+15}{2x^2-9x-5}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\left(x-3\right)\left(x-5\right)}{2x^2-10x+x-5}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\left(x-3\right)\left(x-5\right)}{\left(x-5\right)\left(2x+1\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{x-3}{2x+1}=\dfrac{5-3}{2\cdot5+1}=\dfrac{2}{11}\)

a: \(\lim\limits_{x\rightarrow3}\dfrac{x^2-9}{x^2-5x+6}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x-2\right)}\)

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

b: \(\lim\limits_{x\rightarrow5}\dfrac{x^2-5x}{x-5}=\lim\limits_{x\rightarrow5}\dfrac{x\left(x-5\right)}{x-5}=\lim\limits_{x\rightarrow5}x=5\)

c: \(\lim\limits_{x\rightarrow-3}\dfrac{x^2-3x}{2x^2+9x+9}\)

\(=\lim\limits_{x\rightarrow-3}\dfrac{x\left(x-3\right)}{2x^2+6x+3x+9}\)

\(=\lim\limits_{x\rightarrow-3}\dfrac{\left(-3\right)\left(-3-3\right)}{\left(-3+3\right)\left(2\cdot\left(-3\right)+3\right)}\)

\(=\lim\limits_{x\rightarrow-3}\dfrac{18}{0\cdot\left(-3\right)}=-\infty\)

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

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

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

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

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 \(3-x>0;\forall x< 3\)

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

\(A=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x^3+x^2+x-2\right)}{\left(x-1\right)\left(x^2+x+3\right)}=\lim\limits_{x\rightarrow1}\frac{x^3+x^2+x-2}{x^2+x+3}=\frac{1}{5}\)

\(B=\lim\limits_{x\rightarrow3}\frac{-x^2+2x+3}{\left(x-1\right)\left(x-3\right)\left(\sqrt{2x+3}+x\right)}=\lim\limits_{x\rightarrow3}\frac{\left(3-x\right)\left(x+1\right)}{\left(x-1\right)\left(x-3\right)\left(\sqrt{2x+3}+x\right)}\)

\(=\lim\limits_{x\rightarrow3}\frac{-x-1}{\left(x-1\right)\left(\sqrt{2x+3}+x\right)}=\frac{-4}{2.\left(3+3\right)}=-\frac{1}{3}\)

\(a=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}{\left(x-1\right)\left(x^2+x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x+1\right)\left(x^2+1\right)}{x^2+x-1}=\frac{4}{1}=4\)

\(b=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{x^4-x^3+x^2-x+1}{x^2-x+1}=\frac{5}{3}\)

\(c=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)^2}{\left(x^2+1\right)\left(x^2-9\right)}=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x+3\right)}=\frac{0}{60}=0\)

\(d=\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{x^2-2x+1}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2x-2}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=10\)

\(e=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)

\(f=\lim\limits_{x\rightarrow-2}\frac{\left(x+2\right)\left(x-2\right)\left(x^2+4\right)}{\left(x+2\right)x^2}=\lim\limits_{x\rightarrow-2}\frac{\left(x-2\right)\left(x^2+4\right)}{x^2}=-8\)

Hai câu d, e khai triển thì dài quá nên làm biếng sử dụng L'Hopital

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

a) \(\mathop {\lim }\limits_{x \to 2} \left( {{x^2} - 4x + 3} \right) = \mathop {\lim }\limits_{x \to 2} {x^2} - \mathop {\lim }\limits_{x \to 2} \left( {4x} \right) + 3 = {2^2} - 4.2 + 3 =  - 1\)

b) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 5x + 6}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x - 2} \right)}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \left( {x - 2} \right) = \mathop {\lim }\limits_{x \to 3} x - 2 = 3 - 2 = 1\)

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

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

a: \(=\lim\limits_{x\rightarrow+\infty}\dfrac{4+\dfrac{3}{x}}{2}=\dfrac{4}{2}=2\)

b: \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{2}{x}}{3+\dfrac{1}{x}}=0\)

c: \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x^2}}}{1+\dfrac{1}{x}}=1\)