Bài 2: Giới hạn của hàm số

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

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

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

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

\(=-\infty\) vì \(\left\{{}\begin{matrix}x^2-4>0\\\lim\limits_{x\rightarrow2^+}x^2-4=2^2-4=0\\\lim\limits_{x\rightarrow2^+}-x-1=-2-1=-3< 0\end{matrix}\right.\)

Không tồn tại.

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

\(\lim\limits_{x\rightarrow3^-}\dfrac{\sqrt{x^2-7x+12}}{\sqrt{9-x^2}}\)

\(=\lim\limits_{x\rightarrow3^-}\sqrt{\dfrac{\left(x-3\right)\left(x-4\right)}{\left(3-x\right)\left(3+x\right)}}\)

\(=\lim\limits_{x\rightarrow3^-}\sqrt{\dfrac{4-x}{3+x}}\)

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

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

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

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

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

\(\lim\limits_{x\rightarrow2}\dfrac{2x^3+5x^2-7x+2}{x^2-3x+2}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{2x^3-4x^2+9x^2-18x+11x-22+24}{\left(x-2\right)\left(x+1\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(2x^2+9x+11\right)+24}{\left(x-2\right)\left(x+1\right)}\)

\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}\left(x-2\right)\left(2x^2+9x+11\right)+24=24>0\\\lim\limits_{x\rightarrow2}\left(x-2\right)=2-2=0\\\lim\limits_{x\rightarrow2}x+1=2+1=3>0\end{matrix}\right.\)

\(\lim\limits_{x\rightarrow1}\dfrac{x+2-\sqrt{x+8}}{\sqrt{x+3}+x-3}=\lim\limits_{x\rightarrow1}\left(\dfrac{x^2+3x-4}{x+2+\sqrt{x+8}}:\dfrac{-x^2+7x-6}{\sqrt{x+3}+3-x}\right)\\ =\lim\limits_{x\rightarrow1}\left[\dfrac{\left(x-1\right)\left(x+4\right)}{x+2+\sqrt{x+8}}:\dfrac{\left(x-1\right)\left(6-x\right)}{\sqrt{x+3}+3-x}\right]\\ =\lim\limits_{x\rightarrow1}\dfrac{\left(x+4\right)\left(\sqrt{x+3}+3-x\right)}{\left(6-x\right)\left(x+2+\sqrt{x+8}\right)}=\dfrac{2}{3}\)

\(\lim\limits_{x\rightarrow2}\dfrac{x-\sqrt{3x-2}}{x^4-4}=0\)

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

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

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

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

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

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

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