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

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

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

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

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

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

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

lỗi gõ câu a

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

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

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

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

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

b. \(\lim\limits_{x\rightarrow3^-}\dfrac{x+3}{x-3}=\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}\)

Do \(\lim\limits_{x\rightarrow3^-}\left(-x-3\right)=-6< 0\)

\(\lim\limits_{x\rightarrow3^-}\left(3-x\right)=0\) và \(3-x>0;\forall x< 3\)

\(\Rightarrow\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}=-\infty\)

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: \(\sqrt{x+1} - \sqrt[3]{x-1} = (\sqrt{x+1} - 1) + (1 - \sqrt[3]{x-1})\)

\(\lim_{x\to0}\frac{(x+1) - 1}{x(\sqrt{x+1} + 1)}\)

\(=\lim_{x\to0}\frac{x}{x(\sqrt{x+1} + 1)}\)

\(=\lim_{x\to0}\frac{1}{\sqrt{x+1} + 1}=\frac{1}{1 + 1}=\frac{1}{2}\)

\(\lim_{x\to0}\frac{1 - \sqrt[3]{x-1}}{x}\)

\(=\lim_{x\to0}\frac{2 - x}{x\left(1 + \sqrt[3]{x-1} + \sqrt[3]{(x-1)^2}\right)}\)

mà \(\lim_{x\to0}2-x=2-0=2\) và \(\lim_{x\to0}x\left(1+\sqrt[3]{x-1}+\sqrt[3]{(x-1)^2}\right)=0\)

nên \(\lim_{x\to0}\frac{1 - \sqrt[3]{x-1}}{x}=+\) ∞ khi x->0+; =-∞ khi x->0-

\(\lim_{x \to 0} \frac{\sqrt{x + 1} - \sqrt[3]{x - 1}}{x}\)

=\(\lim_{x\to0}\frac{\sqrt{x+1} - 1}{x}+\lim_{x\to0}\frac{1 - \sqrt[3]{x-1}}{x}\)

=∞

b:

\(\sqrt[3]{x-3} + \sqrt[4]{2x-3} = (\sqrt[3]{x-3} + 1) + (\sqrt[4]{2x-3} - 1)\)

\(\lim_{x\to2}\frac{\sqrt[3]{x-3} + 1}{x - 2}\)

\(=\lim_{x\to2}\frac{(x - 3) + 1}{(x - 2)\left(\sqrt[3]{(x-3)^2} - \sqrt[3]{x-3} + 1\right)}\)

\(=\lim_{x\to2}\frac{x - 2}{(x - 2)\left(\sqrt[3]{(x-3)^2} - \sqrt[3]{x-3} + 1\right)}\)

\(=\lim_{x\to2}\frac{1}{\sqrt[3]{(x-3)^2} - \sqrt[3]{x-3} + 1}=\frac{1}{1 - (-1) + 1}=\frac{1}{3}\)

\(\lim_{x\to2}\frac{\sqrt[4]{2x-3} - 1}{x - 2}\)

\(=\lim_{x\to2}\frac{(2x - 3) - 1}{(x - 2)(\sqrt[4]{2x-3} + 1)(\sqrt{2x-3} + 1)}=\lim_{x\to2}\frac{2(x - 2)}{(x - 2)(\sqrt[4]{2x-3} + 1)(\sqrt{2x-3} + 1)}\)

\(=\lim_{x\to2}\frac{2}{(\sqrt[4]{2x-3} + 1)(\sqrt{2x-3} + 1)}=\frac{2}{(1 + 1)(1 + 1)}=\frac{2}{4}=\frac{1}{2}\)

\(\lim_{x\to2}\frac{\sqrt[3]{x - 3} + \sqrt[4]{2x - 3}}{x - 2}\)

\(=\lim_{x\to2}\frac{\sqrt[3]{x-3} + 1}{x - 2}+\lim_{x\to2}\frac{\sqrt[4]{2x-3} - 1}{x - 2}\)

=1/3+1/2

=5/6

\(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4+4x}-\sqrt[3]{8-x}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4+4x}-2+2-\sqrt[3]{8-x}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x+4-4}{\sqrt{4x+4}+2}+\dfrac{8-8+x}{4+2\cdot\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x}{\sqrt{4x+4}+2}+\dfrac{x}{4+2\cdot\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{4}{\sqrt{4x+4}+2}+\dfrac{1}{4+2\cdot\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}\)

\(=\dfrac{4}{\sqrt{4\cdot0+4}+2}+\dfrac{1}{4+2\cdot\sqrt[3]{8-0}+\sqrt[3]{\left(8-0\right)^2}}\)

\(=\dfrac{4}{2+2}+\dfrac{1}{4+2\cdot2+4}\)

\(=1+\dfrac{1}{12}=\dfrac{13}{12}\)

Đặt \(A=\sqrt{1+2x}\cdot\sqrt[3]{1+3x}-\sqrt{1+4x}\)

=>A'=\(\frac{\sqrt[3]{1+3x}}{\sqrt{1+2x}}+\frac{\sqrt{1+2x}}{\left(1+3x\right)^{\frac23}}-\frac{2}{\sqrt{1+4x}}\)

=>A''\(=\left[-\frac{1}{(1+2x)^{\frac{3}{2}}}\sqrt[3]{1+3x}+\frac{1}{\sqrt{1+2x}}\frac{1}{(1+3x)^{\frac{2}{3}}}\right]+\left[\frac{1}{\sqrt{1+2x}}\frac{1}{(1+3x)^{\frac{2}{3}}}-\frac{2\sqrt{1+2x}}{(1+3x)^{\frac{5}{3}}}\right]+\frac{4}{(1+4x)^{\frac{3}{2}}}\)

Đặt \(B=1+x-\sqrt{1+2x}\)

=>B'=\(1-\frac{2}{2\sqrt{1+2x}}=1-\frac{1}{\sqrt{1+2x}}\)

=>B''=\(\frac{1}{\left(1+2x\right)^{\frac23}}\)

A'(0)=\(\frac{\sqrt[3]{1+3\cdot0}}{\sqrt{1+2\cdot0}}+\frac{\sqrt{1+2\cdot0}}{\left(1+3\cdot0\right)^{\frac23}}-\frac{2}{\sqrt{1+4\cdot0}}=0\)

B'(0)=\(1-\frac{1}{\sqrt{1+2\cdot0}}=0\)

A''(0)\(=(-1+1)+(1-2)+4=3\)

B''(0)=\(\frac{1}{\left(1+2\cdot0\right)^{\frac23}}=1\)

Do đó: \(\lim_{x\to0}\frac{\sqrt{1+2x} \cdot\sqrt[3]{1+3x} - \sqrt{1+4x}}{1+x - \sqrt{1+2x}}=\) A''/B''

=3/1

=3

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

\(=lim_{x->0}1=1\)

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:

\(^{lim}_{x\rightarrow0}\dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+2x}-1}{x}=1+\dfrac{2}{3}=\dfrac{5}{3}\)