Lời giải:

a. \(\lim\limits_{x\to 1+}(x^3+x+1)=3>0\)

\(\lim\limits_{x\to 1+}(x-1)=0\) và $x-1>0$ khi $x>1$

\(\Rightarrow \lim\limits_{x\to 1+}\frac{x^3+x+1}{x-1}=+\infty\)

b.

 \(\lim\limits_{x\to -1+}(3x+2)=-1<0\)

\(\lim\limits_{x\to -1+}(x+1)=0\) và $x+1>0$ khi $x>-1$

\(\Rightarrow \lim\limits_{x\to -1+}\frac{3x+2}{x+1}=-\infty\)

c.

\(\lim\limits_{x\to 2-}(x-15)=-17<0\)

\(\lim\limits_{x\to 2-}(x-2)=0\) và $x-2<0$ khi $x<2$

\(\Rightarrow \lim\limits_{x\to 2-}\frac{x-15}{x-2}=+\infty\)

Em là tám lại ạ

Em là duy khôi ạ

Em là văn tam ạ

Em là mạnh Tuấn ạ

a: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x-2}-\dfrac{12}{x^3-8}\right)\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x+4-12}{\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x-8}{\left(x-2\right)\left(x^2+2x+4\right)}\)

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

\(=\dfrac{2+4}{2^2+2\cdot2+4}=\dfrac{6}{4+4+4}=\dfrac{6}{12}=\dfrac{1}{2}\)

b: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\)

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

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

\(=\lim\limits_{x\rightarrow2}\dfrac{2x-4}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\)

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

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

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

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

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

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

=0

c: \(\lim\limits_{x\rightarrow+\infty}\left[x\cdot\left(\sqrt{x^2+1}-x\right)\right]\)

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

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

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

e: \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{x^2+1-1}{\sqrt{x^2+1}+1}:\dfrac{x^2+16-16}{\sqrt{x^2+16}+4}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+16}+4}{\sqrt{x^2+1}+1}=\dfrac{4+4}{1+1}=\dfrac{8}{2}=4\)

a: \(=lim_{x->-\infty}\dfrac{2x-5+\dfrac{1}{x^2}}{7-\dfrac{1}{x}+\dfrac{4}{x^2}}\)

\(=\dfrac{2x-5}{7}\)

\(=\dfrac{2}{7}x-\dfrac{5}{7}\)

\(=-\infty\)

b: \(=lim_{x->+\infty}x\sqrt{\dfrac{1+\dfrac{1}{x}+\dfrac{3}{x^2}}{3x^2+4-\dfrac{5}{x^2}}}\)

\(=lim_{x->+\infty}x\sqrt{\dfrac{1}{3x^2+4}}=+\infty\)

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

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

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

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

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

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

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

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: \(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x^2+5}-3}{x+2}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{x^2+5-9}{\sqrt{x^2+5}+3}\cdot\dfrac{1}{x+2}\)

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

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

\(=\lim\limits_{x\rightarrow-2}\dfrac{x-2}{\sqrt{x^2+5}+3}\)

\(=\dfrac{-2-2}{\sqrt{\left(-2\right)^2+5}+3}=\dfrac{-4}{3+3}=-\dfrac{4}{6}=-\dfrac{2}{3}\)

b: \(\lim\limits_{x\rightarrow2}\dfrac{x^2+x-6}{x^2-4}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+3x-2x-6}{\left(x-2\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x+3}{x+2}=\dfrac{2+3}{2+2}=\dfrac{5}{4}\)