\(\lim\limits_{x\rightarrow0}\dfrac{\left(1+2013x\right)^{2014}-\left(1-2014x\right)^{2013}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2013.2014\left(1+2013x\right)^{2013}+2013.2014\left(1-2014x\right)^{2012}}{2x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2013^3.2014\left(1+2013x\right)^{2012}-2012.2013.2014^2\left(1-2014x\right)^{2011}}{2}\)

\(=\dfrac{2013^3.2014-2012.2013.2014^2}{2}=...\)

Tui nghĩ cái này L'Hospital chứ giải thông thường là ko ổn :)

\(M=\lim\limits_{x\rightarrow0}\dfrac{\left(1+4x\right)^{\dfrac{1}{2}}-\left(1+6x\right)^{\dfrac{1}{3}}}{1-\cos3x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{2}\left(1+4x\right)^{-\dfrac{1}{2}}.4-\dfrac{1}{3}\left(1+6x\right)^{-\dfrac{2}{3}}.6}{3.\sin3x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-\dfrac{1}{4}.4\left(1+4x\right)^{-\dfrac{3}{2}}.4+\dfrac{2}{9}.6.6\left(1+6x\right)^{-\dfrac{5}{3}}}{3.3.\cos3x}\) 

Giờ thay x vô là được

\(N=\lim\limits_{x\rightarrow0}\dfrac{\left(1+ax\right)^{\dfrac{1}{m}}-\left(1+bx\right)^{\dfrac{1}{n}}}{\left(1+x\right)^{\dfrac{1}{2}}-1}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{m}.\left(1+ax\right)^{\dfrac{1}{m}-1}.a-\dfrac{1}{n}\left(1+bx\right)^{\dfrac{1}{n}-1}.b}{\dfrac{1}{2}\left(1+x\right)^{-\dfrac{1}{2}}}=\dfrac{\dfrac{a}{m}-\dfrac{b}{n}}{\dfrac{1}{2}}\)

\(V=\lim\limits_{x\rightarrow0}\dfrac{\left(1+mx\right)^n-\left(1+nx\right)^m}{\left(1+2x\right)^{\dfrac{1}{2}}-\left(1+3x\right)^{\dfrac{1}{3}}}=\lim\limits_{x\rightarrow0}\dfrac{n\left(1+mx\right)^{n-1}.m-m\left(1+nx\right)^{m-1}.n}{\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{1}{2}}.2-\dfrac{1}{3}\left(1+3x\right)^{-\dfrac{2}{3}}.3}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{n\left(n-1\right)\left(1+mx\right)^{n-2}.m-m\left(m-1\right)\left(1+nx\right)^{m-2}.n}{-\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{3}{2}}.2+\dfrac{2}{9}.3.3\left(1+3x\right)^{-\dfrac{5}{3}}}=....\left(thay-x-vo-la-duoc\right)\)

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: \(\lim\limits_{x->0^-^-}\dfrac{-2x+x}{x\left(x-1\right)}=lim_{x->0^-}\left(\dfrac{-x}{x\left(x-1\right)}\right)\)

\(=lim_{x->0^-}\left(\dfrac{-1}{x-1}\right)=\dfrac{-1}{0-1}=\dfrac{-1}{-1}=1\)

b: \(=lim_{x->-\infty}\left(\dfrac{x^2-x-x^2+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)

\(=lim_{x->-\infty}\left(\dfrac{-x+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)

\(=lim_{x->-\infty}\left(\dfrac{-1+\dfrac{1}{x}}{-\sqrt{1-\dfrac{1}{x^2}}-\sqrt{1-\dfrac{1}{x^2}}}\right)=\dfrac{-1}{-2}=\dfrac{1}{2}\)

1/ \(\lim\limits_{x\rightarrow0^-}\left(\dfrac{x-2}{x^3}\right)=\lim\limits_{x\rightarrow0^-}\dfrac{2-x}{-x^3}=\dfrac{2}{0}=+\infty\)

2/ \(\lim\limits_{x\rightarrow1^+}\dfrac{\left(x^3-x^2\right)^{\dfrac{1}{2}}}{\left(x-1\right)^{\dfrac{1}{2}}+1-x}=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{1}{2}\left(x^3-x^2\right)^{-\dfrac{1}{2}}.\left(3x^2-2x\right)}{\dfrac{1}{2}\left(x-1\right)^{-\dfrac{1}{2}}-1}=0\)

3/ \(\lim\limits_{x\rightarrow1^+}\dfrac{1-\left(x^2+x+1\right)}{x^3-1}=\dfrac{1-3}{0}=-\infty\)

4/ \(\lim\limits_{x\rightarrow-\infty}\left(-\infty-\sqrt[3]{1+\infty}\right)=-\left(\infty+\infty\right)=-\infty?\) Cái này ko chắc :v

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

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

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

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

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

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

=1+1

=2

\(A=\lim\limits_{x\rightarrow0}\dfrac{\left(x^2+2017\right)\left(\sqrt[5]{1-5x}-1\right)+x^2}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{-\dfrac{5x\left(x^2+2017\right)}{\sqrt[5]{\left(1-5x\right)^4}+\sqrt[5]{\left(1-5x\right)^3}+\sqrt[5]{\left(1-5x\right)^2}+\sqrt[5]{1-5x}+1}+x^2}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(-\dfrac{5\left(x^2+2017\right)}{\sqrt[5]{\left(1-5x\right)^4}+\sqrt[5]{\left(1-5x\right)^3}+\sqrt[5]{\left(1-5x\right)^2}+\sqrt[5]{1-5x}+1}+x\right)\)

\(=-2017\)

dễ thấy hàm số trên có dạng 0/0

áp dụng quy tắc l'Hôpital 

\(A=_{\lim\limits_{x\rightarrow0}\dfrac{\left(x^2+2017\right)\sqrt[5]{1-5x}-2017}{x}=\lim\limits_{x\rightarrow0}\dfrac{\left(\left(x^2+2017\right)\sqrt[5]{1-5x}-2017\right)'}{\left(x\right)'}}\)

\(A=\lim\limits_{x\rightarrow0}\dfrac{-x^2-2017}{\sqrt[5]{\left(1-5x\right)^4}}+2x\sqrt[5]{1-5x}=\dfrac{-2017}{1}=-2017\)