Bài 2: Giới hạn của hàm số

\(\lim\limits_{x\rightarrow2}\dfrac{\left(x^2-x-2\right)^2}{x^3-12x+16}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)^2\cdot\left(x+1\right)^2}{x^3-2x^2+2x^2-4x-8x+16}\)

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

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

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x+1\right)^2}{x+4}=\dfrac{\left(2+1\right)^2}{2+4}=\dfrac{9}{6}=\dfrac{3}{2}\)

\(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x+3}-\sqrt{3x+1}}{\sqrt{x-1}}\)

\(=\lim\limits_{x\rightarrow1^+}\left(\dfrac{x+3-3x-1}{\sqrt{x+3}+\sqrt{3x+1}}\cdot\dfrac{1}{\sqrt{x-1}}\right)\)

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

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

\(=\dfrac{-2\cdot\sqrt{1-1}}{\sqrt{1+3}+\sqrt{3\cdot1+1}}=0\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{3x+1}-\sqrt{x+3}}{x^3-1}\)

\(=\lim\limits_{x\rightarrow1}\left(\dfrac{3x+1-x-3}{\sqrt{3x+1}+\sqrt{x+3}}:\left(x-1\right)\left(x^2+x+1\right)\right)\)

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

\(=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(\sqrt{3x+1}+\sqrt{x+3}\right)\left(x^2+x+1\right)}\right)\)

\(=\dfrac{2}{\left(\sqrt{3\cdot1+1}+\sqrt{1+3}\right)\left(1^2+1+1\right)}\)

\(=\dfrac{2}{3\cdot\left(2+2\right)}=\dfrac{2}{3\cdot4}=\dfrac{2}{12}=\dfrac{1}{6}\)

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+4}-2}{3-\sqrt{3x^2+9}}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{x^2+4-4}{\sqrt{x^2+4}+2}:\dfrac{9-3x^2-9}{3+\sqrt{3x^2+9}}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{x^2}{\sqrt{x^2+4}+2}\cdot\dfrac{\sqrt{3x^2+9}+3}{-3x^2}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{3x^2+9}+3}{-3\left(\sqrt{x^2+4}+2\right)}\)

\(=\dfrac{\sqrt{3\cdot0^2+9}+3}{-3\left(\sqrt{0^2+4}+2\right)}=\dfrac{3+3}{-3\left(2+2\right)}=\dfrac{6}{-3\cdot4}=-\dfrac{1}{2}\)

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

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

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

\(=\lim\limits_{x\rightarrow-2}\dfrac{x^2-4}{x^3+5x^2+7x+2}\)

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

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

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

\(\lim\limits_{x\rightarrow1}\dfrac{2x^2-3x+1}{x^3-x^2-x+1}\)

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

\(=\lim\limits_{x\rightarrow1}\dfrac{2x-1}{x^2-1}\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow1}2x-1=2\cdot1-1=1>0\\\lim\limits_{x\rightarrow1}x^2-1=1^2-1=0\end{matrix}\right.\)

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

\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow-3}-x^2-x-6=-\left(-3\right)^2-\left(-3\right)-6=-12< 0\\\lim\limits_{x\rightarrow-3}x^2-9=\left(-3\right)^2-9=0\end{matrix}\right.\)

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

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

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

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

Không tồn tại.

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

\(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow1^-}\sqrt{2-x}=1>0\\\lim\limits_{x\rightarrow1^-}\left(4-x\right)\sqrt{1-x}=0\\x< 1\rightarrow\left(4-x\right)\sqrt{1-x}>0\end{matrix}\right.\\ \rightarrow\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{x^2-3x+2}}{x^2-5x+4}=\lim\limits_{x\rightarrow1^-}\dfrac{\sqrt{2-x}}{\left(4-x\right)\sqrt{1-x}}=+\infty\)