Tính các giới hạn sau:\(M=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{1-cos3x}\)
\(N=\lim\limits_{X\rightarrow0}\dfrac{\sqrt[m]{1+ax}-\sqrt[n]{1+bx}}{\sqrt{1+x}-1}\)
\(V=\lim\limits_{x\rightarrow0}\dfrac{\left(1+mx\right)^n-\left(1+nx\right)^m}{\sqrt{1+2x}-\sqrt[3]{1+3x}}\)
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
\(N=\lim\limits_{x\rightarrow0}\dfrac{\left(1+ax\right)^{\dfrac{1}{m}}-\left(1+bx\right)^{\dfrac{1}{n}}}{\left(1+x\right)^{\dfrac{1}{2}}-1}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{m}.\left(1+ax\right)^{\dfrac{1}{m}-1}.a-\dfrac{1}{n}\left(1+bx\right)^{\dfrac{1}{n}-1}.b}{\dfrac{1}{2}\left(1+x\right)^{-\dfrac{1}{2}}}=\dfrac{\dfrac{a}{m}-\dfrac{b}{n}}{\dfrac{1}{2}}\)
\(V=\lim\limits_{x\rightarrow0}\dfrac{\left(1+mx\right)^n-\left(1+nx\right)^m}{\left(1+2x\right)^{\dfrac{1}{2}}-\left(1+3x\right)^{\dfrac{1}{3}}}=\lim\limits_{x\rightarrow0}\dfrac{n\left(1+mx\right)^{n-1}.m-m\left(1+nx\right)^{m-1}.n}{\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{1}{2}}.2-\dfrac{1}{3}\left(1+3x\right)^{-\dfrac{2}{3}}.3}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{n\left(n-1\right)\left(1+mx\right)^{n-2}.m-m\left(m-1\right)\left(1+nx\right)^{m-2}.n}{-\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{3}{2}}.2+\dfrac{2}{9}.3.3\left(1+3x\right)^{-\dfrac{5}{3}}}=....\left(thay-x-vo-la-duoc\right)\)
Tính các giới hạn
a) \(\lim\limits_{x\rightarrow\infty}\dfrac{a_0x^m+a_1x^{m-1}+a_2x^{m-2}+...+a_m}{b_0x^n+b_1x^{n-1}+b_2x^{n-2}+...+b_n}\)
b) \(\lim\limits_{x\rightarrow+\infty}\dfrac{\left(x-\sqrt{x^2-1}\right)^n+\left(x+\sqrt{x^2-1}\right)^n}{x^n}\)
Tìm các giới hạn sau:
\(\lim\limits_{x\rightarrow-\infty}\) \(\dfrac{\sqrt{x^6+2}}{3\text{x}^3-1}\)
\(\lim\limits_{x\rightarrow+\infty}\) \(\dfrac{\sqrt{x^6+2}}{3\text{x}^3-1}\)
\(\lim\limits_{x\rightarrow-\infty}\) \(\left(\sqrt{2\text{x}^2+1}+x\right)\)
\(\lim\limits_{x\rightarrow1}\) \(\dfrac{2\text{x}^3-5\text{x}-4}{\left(x+1\right)^2}\)
Bài 1. Tìm các giới hạn sau:
a) \(\lim\limits\dfrac{-2n+1}{n}\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}\)
a) \(lim\dfrac{-2n+1}{n}=lim\dfrac{\dfrac{-2n}{n}+\dfrac{1}{n}}{\dfrac{n}{n}}=lim\dfrac{-2+\dfrac{1}{n}}{1}=\dfrac{lim\left(-2\right)+\dfrac{lim1}{n}}{lim1}=\dfrac{-2+0}{1}=-\dfrac{2}{1}=-2\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-\left(x+8\right)}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{3+\sqrt{x+8}}=\dfrac{1}{3+\sqrt{1+8}}=\dfrac{1}{3+3}=\dfrac{1}{9}\)
1/ \(\lim\limits_{x\to 1}\) \(\dfrac{\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1}\)
2/ \(\lim\limits_{x \to \ +\infty} \)\(x\left[\sqrt{4x^2+5}-\sqrt[3]{8x^3-1}\right]\)
3/ \(\lim\limits_{x\to 1}\)\(\dfrac{x^3-2x-1}{x^5-2x-1}\)
Giải giúp mình với ạ
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt[3]{7+x^3}-2\right)-\left(\sqrt{3+x^2}-2\right)}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^3-1}{\left(\sqrt[3]{7+x^3}\right)^2+2\sqrt[3]{7+x^3}+4}-\dfrac{x^2-1}{\sqrt{3+x^2}+2}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^2+x+1}{\left(\sqrt[3]{7+x^3}\right)^2+2\sqrt[3]{7+x^3}+4}-\dfrac{x+1}{\sqrt{3+x^2}+2}}{1}=\dfrac{3}{12}-\dfrac{2}{4}=\dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{4}\).
Tính các giới hạn sau:\(I_1=\lim\limits_{x\rightarrow1}\dfrac{\left(1-\sqrt{x}\right)\left(1-\sqrt[3]{x}\right)....\left(1-\sqrt[n]{x}\right)}{\left(1-x\right)^{n-1}}\)
\(I_2=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+x^2}+x\right)^n-\left(\sqrt{1+x^2}-x\right)^n}{x}\)
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)
\(I_2=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+x^2}+x\right)^n-\left(\sqrt{1+x^2}-x\right)^n}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{n\left(\sqrt{1+x^2}+x\right)^{n-1}\left(\dfrac{x}{\sqrt{1+x^2}}+1\right)-n\left(\sqrt{1+x^2}-x\right)^{n-1}\left(\dfrac{x}{\sqrt{1+x^2}}-1\right)}{1}\)
\(=\dfrac{n.1-n\left(-1\right)}{1}=2n\)
Tính giới hạn
a) \(\lim\limits_{x->0}\dfrac{\sqrt[m]{2x+1}-1}{\sqrt[n]{x+1}-1}\)
b) \(\lim\limits_{x->3}\dfrac{\sqrt[4]{5x+1}-2}{x-3}\)
Tính các giới hạn sau:
a) \(\lim\limits_{x\rightarrow0^-}\dfrac{2\left|x\right|+x}{x^2-x}\)
b) \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\)
c) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{1+x^4+x^6}}{\sqrt{1+x^3+x^4}}\)
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: \(=lim_{x->-\infty}\left(\dfrac{x^2-x-x^2+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)
\(=lim_{x->-\infty}\left(\dfrac{-x+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)
\(=lim_{x->-\infty}\left(\dfrac{-1+\dfrac{1}{x}}{-\sqrt{1-\dfrac{1}{x^2}}-\sqrt{1-\dfrac{1}{x^2}}}\right)=\dfrac{-1}{-2}=\dfrac{1}{2}\)
Tìm các giới hạn sau :
a) \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{4-\sqrt{x^2+16}}\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\)
c) \(\lim\limits_{x\rightarrow+\infty}\dfrac{2x^4+5x-1}{1-x^2+x^4}\)
d) \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+\sqrt{4x^2-x+1}}{1-2x}\)
e) \(\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{x^2+1}-x\right)\)
f) \(\lim\limits_{x\rightarrow2^+}\left(\dfrac{1}{x^2-4}-\dfrac{1}{x-2}\right)\)
