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

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

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

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

a/ \(=\lim\limits_{x\rightarrow-\infty}x^3\left(3+\dfrac{5x^2}{x^3}-\dfrac{9\sqrt{2}x}{x^3}-\dfrac{2017}{x^3}\right)=3.x^3=-\infty\)

b/ \(=\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{1+\dfrac{x}{x^2}+\dfrac{1}{x^2}}-\sqrt[3]{2+\dfrac{x}{x^3}-\dfrac{1}{x^3}}\right)=\left(1-\sqrt[3]{2}\right)x=-\infty\)

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

d/ \(=\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+x^2+1}-x\right)+\lim\limits_{x\rightarrow-\infty}\left(x+\sqrt{x^2+x+1}\right)\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^3+x^2+1-x^3}{\left(\sqrt[3]{x^3+x^2+1}\right)^2+x\sqrt[3]{x^3+x^2+1}-x^2}+\lim\limits_{x\rightarrow-\infty}\dfrac{x^2-x^2-x-1}{x-\sqrt{x^2+x+1}}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+1}{\left(-x\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{x^2}{x^3}+\dfrac{1}{x^3}}\right)^2-x.x\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{x^2}{x^3}+\dfrac{1}{x^3}}-x^2}+\lim\limits_{x\rightarrow-\infty}\dfrac{-x-1}{x+x\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}+\dfrac{1}{x^2}}}\)

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

1. \(=\lim\limits_{x\rightarrow3}\dfrac{\left(3-x\right)\left(3+x\right)\left(\sqrt{x+6}+3\right)}{x-3}=\lim\limits_{x\rightarrow3}-\left(3+x\right)\left(\sqrt{x+6}+3\right)=-36\)

a) \(lim_{x\rightarrow3}\dfrac{9-x^2}{\sqrt{x+6}-3}=lim_{x\rightarrow3}\dfrac{\left(\sqrt{x+6}+3\right)\left(3-x\right)\left(3+x\right)}{\left(\sqrt{x+6}-3\right)\left(\sqrt{x+6}+3\right)}\)

\(=lim_{x\rightarrow3}\dfrac{\left(\sqrt{x+6}+3\right)\left(x-3\right)\left(-x-3\right)}{x+6-9}\)

\(=lim_{x\rightarrow3}\left(\sqrt{x+6}+3\right)\left(-x-3\right)\)

\(=\left(\sqrt{3+6}+3\right)\left(-3-3\right)=-36\)

b) \(lim_{x\rightarrow-\infty}\left(\sqrt{x^2+x}+x\right)=lim_{x\rightarrow-\infty}\dfrac{\left(\sqrt{x^2+x}+x\right)\left(\sqrt{x^2+x}-x\right)}{\sqrt{x^2+x}-x}\)

\(=lim_{x\rightarrow-\infty}\dfrac{x}{\sqrt{x^2+x}-x}=lim_{x\rightarrow-\infty}\dfrac{1}{\sqrt{1+\dfrac{1}{x}}-1}=\dfrac{-1}{2}\)

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: \(\lim\limits_{x\rightarrow-2}\dfrac{4-x^2}{2x^2+7x+6}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{\left(2-x\right)\left(2+x\right)}{2x^2+4x+3x+6}\)

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

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

b: \(\lim\limits_{x\rightarrow4}\dfrac{2x^2-13x+20}{x^3+64}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{2x^2-8x-5x+20}{\left(x+4\right)\left(x^2-4x+16\right)}\)

\(=\lim\limits_{x\rightarrow4}\dfrac{\left(x-4\right)\left(2x-5\right)}{x^3+64}\)

\(=\dfrac{\left(4-4\right)\left(2\cdot4-5\right)}{4^3+64}=0\)

c: \(\lim\limits_{x\rightarrow-1}\dfrac{2x^2+8x+6}{-2x^2+7x+9}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{2x^2+2x+6x+6}{-2x^2-2x+9x+9}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(2x+6\right)}{-2x\left(x+1\right)+9\left(x+1\right)}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(2x+6\right)}{\left(x+1\right)\left(-2x+9\right)}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{2x+6}{-2x+9}=\dfrac{2\cdot\left(-1\right)+6}{-2\cdot\left(-1\right)+9}\)

\(=\dfrac{4}{11}\)

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

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

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

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

\(A=\lim\limits_{x\rightarrow2}\frac{\left(x-2\right)\left(2x-1\right)}{x-2}=\lim\limits_{x\rightarrow2}\left(2x-1\right)=3\)

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

\(C=\lim\limits_{x\rightarrow2}\frac{x^2+2x}{x^2+4x+4}=\frac{4+4}{4+8+4}=\frac{1}{2}\)

\(D=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x^2-1\right)}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow1}\frac{x^2-1}{x-2}=\frac{0}{-1}=0\)

\(E=\lim\limits_{x\rightarrow1}\frac{x^3-5x^2+3x+9}{x^4-8x^4-9}=\frac{1-5+3+9}{1-8-9}=-\frac{1}{2}\)

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

\(G=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(2x+1\right)}=\lim\limits_{x\rightarrow1}\frac{x+3}{2x+1}=\frac{4}{3}\)

\(H=\lim\limits_{x\rightarrow-2}\frac{\left(x+2\right)\left(x-1\right)^2}{\left(2-x\right)\left(x+2\right)}=\lim\limits_{x\rightarrow-2}\frac{\left(x-1\right)^2}{2-x}=\frac{9}{4}\)

\(I=\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+1}{x^2-1}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4}{2x}=\frac{24-25}{2}=-\frac{1}{2}\)

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

\(\lim\limits_{x\rightarrow6}\dfrac{f\left(x\right)-6}{x-6}\) hữu hạn \(\Rightarrow f\left(6\right)=6\)

\(...=\lim\limits_{x\rightarrow6}\dfrac{\dfrac{f\left(x\right)-6}{\sqrt[3]{\left[f\left(x\right)+21\right]^2}+3\sqrt[3]{f\left(x\right)+21}+9}}{x-6}\)

\(=\dfrac{9}{2}.\dfrac{1}{\sqrt[3]{\left(6+21\right)^2}+3\sqrt[3]{6+21}+9}\)