Bài 2: Giới hạn của hàm số

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

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

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

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

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

\(=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\)

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+1}+\sqrt{x+4}-3}{x}\)

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

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

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

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sqrt{x+1}+1}+\dfrac{1}{\sqrt{x+4}+2}\right)\)

\(=\dfrac{1}{\sqrt{0+1}+1}+\dfrac{1}{\sqrt{0+4}+2}=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4}\)

$x\to -\infty$ thì sao tồn tại $\sqrt{x}$ được nhỉ. Bạn xem lại đề.

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

\(\lim\limits_{x\rightarrow-\infty}\left(1-3x-\sqrt{9x^2-2x+1}\right)\)

\(=\lim\limits_{x\rightarrow-\infty}\left(1-\dfrac{9x^2-9x^2+2x-1}{3x-\sqrt{9x^2-2x+1}}\right)\)

\(=\lim\limits_{x\rightarrow-\infty}\left(1-\dfrac{2x-1}{3x-\sqrt{9x^2-2x+1}}\right)\)

\(=\lim\limits_{x\rightarrow-\infty}\left(1-\dfrac{2-\dfrac{1}{x}}{3+\sqrt{9-\dfrac{2}{x}+\dfrac{1}{x^2}}}\right)\)

\(=1-\dfrac{2-0}{3+\sqrt{9-0+0}}=1-\dfrac{2}{3+3}=1-\dfrac{1}{3}=\dfrac{2}{3}\)

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

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

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

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

\(\lim\limits_{x\rightarrow5^-}\left(\sqrt{5-x}+2x\right)\)

\(=\sqrt{5-5}+2\cdot5\)

=10

\(\lim\limits_{x\rightarrow3^-}\sqrt{x^2+8x+3}\)

\(=\sqrt{\left(-3\right)^2+8\cdot\left(-3\right)^2+3}\)

\(=\sqrt{9+8\cdot9+3}=\sqrt{12+72}=\sqrt{84}=2\sqrt{21}\)

\(\lim\limits_{x\rightarrow3^-}\sqrt{x^2+8x+3}=\sqrt{3^2+8.3+3}=6\)

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

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

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

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