a/ L'Hospital:

 \(=\lim\limits_{x\rightarrow2}\dfrac{x-\left(x+2\right)^{\dfrac{1}{2}}}{\left(4x+1\right)^{\dfrac{1}{2}}-3}=\lim\limits_{x\rightarrow2}\dfrac{1-\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}{\dfrac{1}{2}\left(4x+1\right)^{-\dfrac{1}{2}}.4}=\dfrac{1-\dfrac{1}{2}.4^{-\dfrac{1}{2}}}{2.9^{-\dfrac{1}{2}}}=\dfrac{9}{8}\)

b/ L'Hospital:\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7\right)^{\dfrac{1}{2}}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{2}\left(2x+7\right)^{-\dfrac{1}{2}}.2+1}{3x^2-8x}=\dfrac{9^{-\dfrac{1}{2}}+1}{3-8}=-\dfrac{4}{15}\)

Kiểm tra lại đề bài, giới hạn này không tồn tại

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

\(\left(...\right)=\lim\limits_{x\rightarrow1}\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(\sqrt{2x+7}+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{2}{\sqrt{2x+7}+3}=\dfrac{1}{3}\)

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

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

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

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

\(=\dfrac{\left(1+1\right)\left(\sqrt{3+1}+2\right)}{2}=\dfrac{2\cdot4}{3}=\dfrac{8}{3}\)

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

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

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

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

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

\(=2\cdot4=8\)

Chúng ta tính giới hạn sau:

\(\lim\limits_{x\rightarrow1}\dfrac{1-\sqrt[n]{x}}{1-x}\)

Cách đơn giản nhất là sử dụng L'Hopital:

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

Phức tạp hơn thì tách mẫu theo hằng đẳng thức

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

Tóm lại ta có:

\(\lim\limits_{x\rightarrow1}\dfrac{1-\sqrt[n]{x}}{1-x}=\dfrac{1}{n}\)

Do đó:

\(I_1=\lim\limits_{x\rightarrow1}\left(\dfrac{1-\sqrt[2]{x}}{1-x}\right)\left(\dfrac{1-\sqrt[3]{x}}{1-x}\right)...\left(\dfrac{1-\sqrt[n]{x}}{1-x}\right)=\dfrac{1}{2}.\dfrac{1}{3}...\dfrac{1}{n}=\dfrac{1}{n!}\)

Câu 2 cũng vậy: L'Hopital hoặc tách hằng đẳng thức trâu bò (thôi L'Hopital đi cho đỡ sợ)

\(I_2=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+x^2}+x\right)^n-\left(\sqrt{1+x^2}-x\right)^n}{x}\)

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

\(=\dfrac{n.1-n\left(-1\right)}{1}=2n\)

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

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

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

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

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

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

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

\(=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)