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a. Áp dụng công thức L'Hospital:
\(\lim\limits_{x\to 0}\frac{\sqrt{x+1}-\sqrt{1-x}}{\sqrt[3]{x+1}-\sqrt{1-x}}=\lim\limits_{x\to 0}\frac{\frac{1}{2}(x+1)^{\frac{-1}{2}}+\frac{1}{2}(1-x)^{\frac{-1}{2}}}{\frac{1}{3}(x+1)^{\frac{-2}{3}}+\frac{1}{2}(1-x)^{\frac{-1}{2}}}=\frac{1}{\frac{5}{6}}=\frac{6}{5}\)
b.
\(\lim\limits_{x\to 0}(\frac{1}{x}-\frac{1}{x^2})=\lim\limits_{x\to 0}\frac{x-1}{x^2}=-\infty\)
c. Áp dụng quy tắc L'Hospital:
\(\lim\limits_{x\to +\infty}\frac{x^4-x^3+11}{2x-7}=\lim\limits_{x\to +\infty}\frac{4x^3-3x^2}{2}=+\infty \)
d.
\(\lim\limits_{x\to 5}\frac{7}{(x-1)^2}.\frac{2x+1}{2x-3}=\frac{7}{(5-1)^2}.\frac{2.5+11}{2.5-3}=\frac{11}{16}\)
\(lim_{x\rightarrow1}\frac{x^3+2x-3}{x^2-x}\)
\(=lim_{x\rightarrow1}\frac{\left(x-1\right)\left(x^2+x+3\right)}{x\left(x-1\right)}\)
\(=lim_{x\rightarrow1}\frac{x^2+x+3}{x}\)
\(=\frac{1^2+1+3}{1}\)
\(=5\)
\(lim_{x\rightarrow1}\frac{\sqrt{2x+2}-\sqrt{3x+1}}{x-1}\)
\(=lim_{x\rightarrow1}\frac{\left(2x+2\right)-\left(3x+1\right)}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)
\(=lim_{x\rightarrow1}\frac{2x+2-3x-1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)
\(=lim_{x\rightarrow1}\frac{-x+1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)
\(=lim_{x\rightarrow1}\frac{-1\left(x-1\right)}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)
\(=lim_{x\rightarrow1}\frac{-1}{\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)
\(=\frac{-1}{\sqrt{2\cdot1+2}+\sqrt{3\cdot1+1}}\)
\(=\frac{-1}{2+2}=\frac{-1}{4}\)
lim ( x ----> 0 ) \(\frac{\sqrt[m]{1+ax}-\sqrt[n]{1+bx}}{x}\)
= lim ( x----> 0 ) \(\frac{\sqrt[m]{1+ax}-1+1-\sqrt[n]{1+bx}}{x}\)
= lim ( x ---> 0 ) \(\frac{\sqrt[m]{1+ax}-1}{x}\)- lim ( x ---> 0 ) \(\frac{\sqrt[n]{1+bx}-1}{x}\)
= lim ( x ----> 0 ) \(\frac{ax}{x\left(\sqrt[m]{\left(1+ax\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1\right)}\)
- lim ( x ----> 0 ) \(\frac{bx}{x\left(\sqrt[n]{\left(1+ax\right)^{n-1}}+\sqrt[n]{\left(1+ax\right)^{n-2}}+...+1\right)}\)
= lim ( x -----> 0 ) \(\frac{a}{\sqrt[m]{\left(1+ax\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1}\)
- lim ( x ---> 0 ) \(\frac{b}{\sqrt[n]{\left(1+bx\right)^{n-1}}+\sqrt[n]{\left(1+bx\right)^{n-2}}+...+1}\)
= \(\frac{a}{m}-\frac{b}{n}\)
\(\lim\limits_{x\rightarrow0}\frac{\left(2x+1\right)^{\frac{1}{2}}-\left(3x+1\right)^{\frac{1}{3}}}{x^2}=\lim\limits_{x\rightarrow0}\frac{\left(2x+1\right)^{-\frac{1}{2}}-\left(3x+1\right)^{-\frac{2}{3}}}{2x}\)
\(=\lim\limits_{x\rightarrow0}\frac{-\left(2x+1\right)^{-\frac{3}{2}}+2\left(3x+1\right)^{-\frac{5}{3}}}{2}=\frac{-1+2}{2}=\frac{1}{2}\)
Ta xét:
\(\sqrt{1+2x}\cdot\sqrt[3]{1+3x}-1\)
\(=\sqrt{1+2x}-\sqrt{1+2x}+\sqrt{1+2x}\cdot\sqrt[3]{1+2x}-1\)
\(=\left(\sqrt{1+2x}-1\right)+\sqrt{1+2x}\cdot\left(\sqrt[3]{1+2x}-1\right)\)
Xét giới hạn trên:
\(\Rightarrow^{lim}_{x\rightarrow0}\dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+2x}-1}{x}\)
\(=^{lim}_{x\rightarrow0}\left(\dfrac{\sqrt{1+2x}-1}{x}\right)+^{lim}_{x\rightarrow0}\left(\dfrac{\sqrt{1+2x}\cdot\left(\sqrt[3]{1+2x}-1\right)}{3}\right)\)
Tính giới hạn từng thành phần:
* \(^{lim}_{x\rightarrow0}\left(\dfrac{\sqrt{1+2x}-1}{x}\right)=^{lim}_{x\rightarrow0}\left(\dfrac{1+2x-1}{x\left(\sqrt{1+2x}+1\right)}\right)\)
\(=^{lim}_{x\rightarrow0}\left(\dfrac{2}{\sqrt{1+2x}+1}\right)=\dfrac{2}{\sqrt{1+2\cdot0}+1}=1\left(1\right)\)
* \(^{lim}_{x\rightarrow0}\left(\dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+2x}-1}{x}\right)\)
\(=^{lim}_{x\rightarrow0}\left(\sqrt{1+2x}\cdot\dfrac{1+2x-1}{x\left(\left(\sqrt[3]{1+2x}\right)^2+\sqrt[3]{1+2x}+1\right)}\right)\)
\(=^{lim}_{x\rightarrow0}\left(\sqrt{1+2x}\cdot\dfrac{2}{\left(\sqrt[3]{1+2x}\right)^2+\sqrt[3]{1+2x}+1}\right)\)
\(=\sqrt{1+2\cdot0}\cdot\dfrac{2}{(\sqrt[3]{1+2\cdot0})^2+\sqrt[3]{1+2\cdot0}+1}\)
\(=\dfrac{2}{3}\left(2\right)\)
Lấy \(\left(1\right)+\left(2\right)\) ta được:
\(^{lim}_{x\rightarrow0}\dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+2x}-1}{x}=1+\dfrac{2}{3}=\dfrac{5}{3}\)
\(=\frac{\left|x\right|\sqrt{1+\frac{2}{x}}+3x}{\left|x\right|\sqrt{4+\frac{1}{x^2}}-x+3}=\frac{-x\left(\sqrt{1+\frac{2}{x}}-3\right)}{-x\left(\sqrt{4+\frac{1}{x^2}}+1+\frac{3}{x}\right)}=\frac{1-3}{2+1+0}=...\)