a: \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x}-\sqrt[3]{x^2+7}}{x^2-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x}-2+2-\sqrt[3]{x^2+7}}{x^2-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{5-x-4}{\sqrt{5-x}+2}+\dfrac{8-x^2-7}{4+2\sqrt[3]{x^2+7}+\sqrt[3]{\left(x^2+7\right)^2}}}{x^2-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1-x}{\sqrt{5-x}+2}+\dfrac{1-x^2}{4+2\sqrt[3]{x^2+7}+\sqrt[3]{\left(x^2+7\right)^2}}}{x^2-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(1-x\right)\left(\dfrac{1}{\sqrt{5-x}+2}+\dfrac{1+x}{4+2\sqrt[3]{x^2+7}+\sqrt[3]{\left(x^2+7\right)^2}}\right)}{-\left(1-x\right)\left(1+x\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{\sqrt{5-x}+2}+\dfrac{1+x}{4+2\sqrt[3]{x^2+7}+\sqrt[3]{\left(x^2+7\right)^2}}}{-\left(1+x\right)}\)

\(=\dfrac{\dfrac{1}{\sqrt{5-1}+2}+\dfrac{1+1}{4+2\cdot\sqrt[3]{1^2+7}+\sqrt[3]{\left(1+7\right)^2}}}{-\left(1+1\right)}\)

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

\(=\dfrac{\dfrac{1}{3}+\dfrac{1}{6}}{-2}=-\dfrac{1}{4}\)

b: \(\lim\limits_{x\rightarrow4}\dfrac{x^2-4x}{x^2+x-20}\)

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

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

\(=\lim\limits_{x\rightarrow4}\dfrac{x}{x+5}=\dfrac{4}{4+5}=\dfrac{4}{9}\)

a: \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-\sqrt{5-x^2}}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-2+2-\sqrt{5-x^2}}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x+7-8}{\sqrt[3]{\left(x+7\right)^2}+2\cdot\sqrt[3]{x+7}+4}+\dfrac{4-5+x^2}{2+\sqrt{5-x^2}}}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\cdot\sqrt[3]{x+7}+4}+\dfrac{x^2-1}{2+\sqrt{5-x^2}}}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(\dfrac{1}{\sqrt[3]{\left(x+7\right)^2}+2\cdot\sqrt[3]{\left(x+7\right)}+4}+\dfrac{x+1}{2+\sqrt{5-x^2}}\right)}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{1}{\sqrt[3]{\left(x+7\right)^2}+2\cdot\sqrt[3]{x+7}+4}+\dfrac{x+1}{2+\sqrt{5-x^2}}\)

\(=\dfrac{1}{\sqrt[3]{\left(1+7\right)^2}+2\cdot\sqrt[3]{1+7}+4}+\dfrac{1+1}{2+\sqrt{5-1^2}}\)

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

\(=\dfrac{1}{12}+\dfrac{1}{2}=\dfrac{7}{12}\)

b: \(\lim\limits_{x\rightarrow5}\dfrac{x-5}{\sqrt{x}-\sqrt{5}}\)

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

\(=\lim\limits_{x\rightarrow5}\sqrt{x}+\sqrt{5}=\sqrt{5}+\sqrt{5}=2\sqrt{5}\)

a: \(\lim\limits_{x\rightarrow+\infty}\dfrac{x-2}{3-\sqrt{x^2+7}}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1-\dfrac{2}{x}}{\dfrac{3}{x}-\sqrt{1+\dfrac{7}{x^2}}}\)

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

b: \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2-x}-\sqrt{4x^2+1}}{2x+3}\)

\(=\dfrac{\sqrt{x^2\left(1-\dfrac{1}{x}\right)}-\sqrt{x^2\left(4+\dfrac{1}{x^2}\right)}}{2x+3}\)

\(=\dfrac{-x\cdot\sqrt{1-\dfrac{1}{x}}+x\cdot\sqrt{4+\dfrac{1}{x^2}}}{x\left(2+\dfrac{3}{x}\right)}\)

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

\(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt[3]{2x+12}+x}{x^2+2x}\)

\(=\lim\limits_{x\rightarrow-2}\left(\dfrac{2x+12+x^3}{\sqrt[3]{\left(2x+12\right)^2}-x\cdot\sqrt[3]{2x+12}+x^2}\cdot\dfrac{1}{x^2+2x}\right)\)

\(=\lim\limits_{x\rightarrow-2}\left(\dfrac{x^3+2x^2-2x^2-4x+6x+12}{\left(\sqrt[3]{\left(2x+12\right)^2}+x\cdot\sqrt[3]{2x+12}+x^2\right)\cdot x\cdot\left(x+2\right)}\right)\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{\left(x+2\right)\left(x^2-2x+6\right)}{\left(\sqrt[3]{\left(2x+12\right)^2}-x\cdot\sqrt[3]{2x+12}+x^2\right)\cdot x\cdot\left(x+2\right)}\)

\(=\lim\limits_{x\rightarrow-2}\dfrac{x^2-2x+6}{x\cdot\left(\sqrt[3]{\left(2x+12\right)^2}-x\cdot\sqrt[3]{2x+12}+x^2\right)}\)

\(=\dfrac{\left(-2\right)^2-2\cdot\left(-2\right)+6}{\sqrt[3]{\left(-2\cdot2+12\right)^2}-\left(-2\right)\cdot\sqrt[3]{2\cdot\left(-2\right)+12}+\left(-2\right)^2}\)

\(=\dfrac{4+4+6}{\sqrt[3]{64}+2\cdot\sqrt[3]{8}+4}\)

\(=\dfrac{14}{8+2\cdot2+4}=\dfrac{14}{12+4}=\dfrac{14}{16}=\dfrac{7}{8}\)

b: \(\lim\limits_{x\rightarrow2}\dfrac{x-\sqrt{x+2}}{x^3-8}\)

\(=\lim\limits_{x\rightarrow2}\left(\dfrac{x^2-x-2}{x+\sqrt{x+2}}\cdot\dfrac{1}{x^3-8}\right)\)

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

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

\(=\dfrac{2+1}{\left(2+\sqrt{2+2}\right)\left(2^2+2\cdot2+4\right)}\)

\(=\dfrac{3}{\left(2+2\right)\left(4+4+4\right)}=\dfrac{3}{12\cdot4}=\dfrac{1}{4\cdot4}=\dfrac{1}{16}\)

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

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

\(=\lim\limits_{x\rightarrow2}\dfrac{\sqrt{2^2+3}-1}{2^2-3\cdot2+2}\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}\sqrt{2^2+3}-1=\sqrt{7}-1>0\\\lim\limits_{x\rightarrow2}2^2-3\cdot2+2=0\end{matrix}\right.\)

b: \(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{4x-1}+3}{x^2-4}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{4x-1-9}{\sqrt{4x-1}-3}\cdot\dfrac{1}{x^2-4}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{4x-10}{\sqrt{4x-1}-3}\cdot\dfrac{1}{\left(x-2\right)\left(x+2\right)}\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}\dfrac{4x-10}{\sqrt{4x-1}-3}=\dfrac{4\cdot2-10}{\sqrt{4\cdot2-1}-3}=\dfrac{-2}{\sqrt{7}-3}>0\\\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x-2\right)\cdot\left(x+2\right)}=\dfrac{1}{\left(2+2\right)\cdot\left(2-2\right)}=+\infty\end{matrix}\right.\)

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

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

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

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

1/ \(\lim\limits_{x\rightarrow0^-}\left(\dfrac{x-2}{x^3}\right)=\lim\limits_{x\rightarrow0^-}\dfrac{2-x}{-x^3}=\dfrac{2}{0}=+\infty\)

2/ \(\lim\limits_{x\rightarrow1^+}\dfrac{\left(x^3-x^2\right)^{\dfrac{1}{2}}}{\left(x-1\right)^{\dfrac{1}{2}}+1-x}=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{1}{2}\left(x^3-x^2\right)^{-\dfrac{1}{2}}.\left(3x^2-2x\right)}{\dfrac{1}{2}\left(x-1\right)^{-\dfrac{1}{2}}-1}=0\)

3/ \(\lim\limits_{x\rightarrow1^+}\dfrac{1-\left(x^2+x+1\right)}{x^3-1}=\dfrac{1-3}{0}=-\infty\)

4/ \(\lim\limits_{x\rightarrow-\infty}\left(-\infty-\sqrt[3]{1+\infty}\right)=-\left(\infty+\infty\right)=-\infty?\) Cái này ko chắc :v

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 $x<2$

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

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

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

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

b: \(\lim\limits_{x\rightarrow-\infty}\dfrac{3x^2-4x+6}{x-2}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2\left(3-\dfrac{4}{x}+\dfrac{6}{x^2}\right)}{x\left(1-\dfrac{2}{x}\right)}\)

\(=\lim\limits_{x\rightarrow-\infty}\left[x\cdot\dfrac{3-\dfrac{4}{x}+\dfrac{6}{x^2}}{1-\dfrac{2}{x}}\right]\)

\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow-\infty}x=-\infty\\\lim\limits_{x\rightarrow-\infty}\dfrac{3-\dfrac{4}{x}+\dfrac{6}{x^2}}{1-\dfrac{2}{x}}=\dfrac{3-0+0}{1-0}=\dfrac{3}{1}=3>0\end{matrix}\right.\)