a) \(\lim\limits_{x\rightarrow-2}\dfrac{2x^2+x-6}{x^3+8}=\lim\limits_{x\rightarrow-2}\dfrac{\left(2x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\\ =\lim\limits_{x\rightarrow-2}\dfrac{2x-3}{x^2-2x+4}=-\dfrac{7}{12}\).

b) \(\lim\limits_{x\rightarrow3}\dfrac{x^4-x^2-72}{x^2-2x-3}=\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\\ =\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)}{x+1}=\dfrac{51}{2}\).

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

d) \(\lim\limits_{x\rightarrow1}\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\\ =\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{x+1}=-\dfrac{1}{2}\).

em cảm ơn ạ !

a) \(lim\dfrac{-2n+1}{n}=lim\dfrac{\dfrac{-2n}{n}+\dfrac{1}{n}}{\dfrac{n}{n}}=lim\dfrac{-2+\dfrac{1}{n}}{1}=\dfrac{lim\left(-2\right)+\dfrac{lim1}{n}}{lim1}=\dfrac{-2+0}{1}=-\dfrac{2}{1}=-2\)

b) \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-\left(x+8\right)}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{3+\sqrt{x+8}}=\dfrac{1}{3+\sqrt{1+8}}=\dfrac{1}{3+3}=\dfrac{1}{9}\)

a) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{6x + 8}}{{5x - 2}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{x\left( {6 + \frac{8}{x}} \right)}}{{x\left( {5 - \frac{2}{x}} \right)}} = \frac{6}{5}\)

b) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{6x + 8}}{{5x - 2}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\left( {6 + \frac{8}{x}} \right)}}{{x\left( {5 - \frac{2}{x}} \right)}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{6 + \frac{8}{x}}}{{5 - \frac{2}{x}}} = \frac{6}{5}\).

c) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {9{x^2} - x + 1} }}{{3x - 2}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x\sqrt {9 - \frac{1}{x} + \frac{1}{{{x^2}}}} }}{{x\left( {3 - \frac{2}{x}} \right)}} =  - \frac{3}{3} =  - 1\).

d) \(\mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {9{x^2} - x + 1} }}{{3x - 2}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{x\sqrt {9 - \frac{1}{x} + \frac{1}{{{x^2}}}} }}{{x\left( {3 - \frac{2}{x}} \right)}} = \frac{3}{3} = 1\).

e) \(\mathop {\lim }\limits_{x \to  - {2^ - }} \frac{{3{x^2} + 4}}{{2x + 4}} =  - \infty \)

Do \(\mathop {\lim }\limits_{x \to  - {2^ - }} \left( {3{x^2} + 1} \right) = 3.{\left( { - 2} \right)^2} + 1 = 13 > 0\) và \(\mathop {\lim }\limits_{x \to  - {2^ - }} \frac{1}{{2x + 4}} =  - \infty \)

g) \(\mathop {\lim }\limits_{x \to  - {2^ + }} \frac{{3{x^2} + 4}}{{2x + 4}} =  + \infty \).

Do \(\mathop {\lim }\limits_{x \to  - {2^ + }} \left( {3{x^2} + 1} \right) = 3.{\left( { - 2} \right)^2} + 1 = 13 > 0\) và \(\mathop {\lim }\limits_{x \to  - {2^ + }} \frac{1}{{2x + 4}} =  + \infty \)

\(=\lim\limits_{x\rightarrow0}\dfrac{2\left(\sqrt[]{2x+1}-1\right)+2-\sqrt[3]{x^2+x+8}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2.2x}{\sqrt[]{2x+1}+1}-\dfrac{x\left(x+1\right)}{\sqrt[3]{\left(x^2+x+8\right)^2}+2\sqrt[3]{x^2+x+8}+4}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{2x+1}+1}-\dfrac{x+1}{\sqrt[3]{\left(x^2+x+8\right)^2}+2\sqrt[3]{x^2+x+8}+4}\right)\)

\(=\dfrac{23}{12}\)

Đề bị lỗi công thức rồi bạn. Bạn cần viết lại để được hỗ trợ tốt hơn.

\(\lim\limits_{x\rightarrow8}\dfrac{x^2-8x}{2-\sqrt[3]{x}}=\lim\limits_{x\rightarrow8}\dfrac{x\left(x-8\right)\left(4+2\sqrt[3]{x}+\sqrt[3]{x^2}\right)}{8-x}\)

\(=\lim\limits_{x\rightarrow8}-x\left(4+2\sqrt[3]{x}+\sqrt[3]{x^2}\right)\)

\(=-8\left(4+2\sqrt[3]{8}+\sqrt[3]{8^2}\right)=-96\)

\(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4+4x}-\sqrt[3]{8-x}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4+4x}-2+2-\sqrt[3]{8-x}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x+4-4}{\sqrt{4x+4}+2}+\dfrac{8-8+x}{4+2\cdot\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x}{\sqrt{4x+4}+2}+\dfrac{x}{4+2\cdot\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{4}{\sqrt{4x+4}+2}+\dfrac{1}{4+2\cdot\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}\)

\(=\dfrac{4}{\sqrt{4\cdot0+4}+2}+\dfrac{1}{4+2\cdot\sqrt[3]{8-0}+\sqrt[3]{\left(8-0\right)^2}}\)

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

\(=1+\dfrac{1}{12}=\dfrac{13}{12}\)

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

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

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

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

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

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

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

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

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 \(3-x>0;\forall x< 3\)

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