\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+3x^2}-\sqrt{x^2-2x}\right)\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+2x}.\sqrt[3]{1+4x}-1}{x}\)
\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+3x^2}-\sqrt{x^2-2x}\right)\)
\(=\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+3x^2}-x+x-\sqrt{x^2-2x}\right)\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^3+3x^2-x^3}{\sqrt[3]{\left(x^3+3x^2\right)^3}+x\cdot\sqrt[3]{x^3+3x^2}+x^2}+\dfrac{x^2-x^2+2x}{x+\sqrt{x^2-2x}}\)
\(=\lim\limits_{x\rightarrow-\infty}\left(\dfrac{3x^2}{\sqrt[3]{\left(x^3+3x^2\right)^3}+x\cdot\sqrt[3]{x^3+3x^2}+x^2}+\dfrac{2x}{x+\sqrt{x^2-2x}}\right)\)
\(=\lim\limits_{x\rightarrow-\infty}\left(\dfrac{3}{\sqrt[3]{\left(1+\dfrac{3}{x}\right)^3}+\sqrt[3]{1+\dfrac{3}{x}}+\dfrac{1}{x}}+\dfrac{2}{1+\sqrt{1-\dfrac{2}{x}}}\right)\)
\(=\dfrac{3}{1+1+1}+\dfrac{2}{1+1}\)
=1+1
=2
\(I=\lim\limits_{x\rightarrow0}\dfrac{^{ }e^{\sin^2\left(ax\right)}\left(1-\sqrt{1+x^2}\right)}{x^2\tan\left(bx\right)}\)
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
\(N=\lim\limits_{x\rightarrow0}\dfrac{\left(1+ax\right)^{\dfrac{1}{m}}-\left(1+bx\right)^{\dfrac{1}{n}}}{\left(1+x\right)^{\dfrac{1}{2}}-1}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{m}.\left(1+ax\right)^{\dfrac{1}{m}-1}.a-\dfrac{1}{n}\left(1+bx\right)^{\dfrac{1}{n}-1}.b}{\dfrac{1}{2}\left(1+x\right)^{-\dfrac{1}{2}}}=\dfrac{\dfrac{a}{m}-\dfrac{b}{n}}{\dfrac{1}{2}}\)
\(V=\lim\limits_{x\rightarrow0}\dfrac{\left(1+mx\right)^n-\left(1+nx\right)^m}{\left(1+2x\right)^{\dfrac{1}{2}}-\left(1+3x\right)^{\dfrac{1}{3}}}=\lim\limits_{x\rightarrow0}\dfrac{n\left(1+mx\right)^{n-1}.m-m\left(1+nx\right)^{m-1}.n}{\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{1}{2}}.2-\dfrac{1}{3}\left(1+3x\right)^{-\dfrac{2}{3}}.3}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{n\left(n-1\right)\left(1+mx\right)^{n-2}.m-m\left(m-1\right)\left(1+nx\right)^{m-2}.n}{-\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{3}{2}}.2+\dfrac{2}{9}.3.3\left(1+3x\right)^{-\dfrac{5}{3}}}=....\left(thay-x-vo-la-duoc\right)\)
Các bạn tính giúp mình mấy câu này với:
1. \(\lim\limits_{x\rightarrow\left(-1\right)-}\dfrac{\sqrt{x^2-3x-4}}{1-x^2}\)
2. \(\lim\limits_{x\rightarrow2^+}\left(\dfrac{1}{x-2}-\dfrac{x+1}{\sqrt{x+2}-2}\right)\)
3. \(\lim\limits_{x\rightarrow+\infty}\dfrac{3x^2-5sin2x+7cos^2x}{2x^2+2}\)
4. \(\lim\limits_{x\rightarrow+\infty}\left(x.sin\left(\dfrac{1}{3x}\right)\right)\)
5. \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{2x+1}.\sqrt[3]{3x+1}.\sqrt[4]{4x+1}-1}{x}\)
6. \(\lim\limits_{x\rightarrow0}\left(\dfrac{\sqrt{9x+4}-\sqrt[3]{4x^{^2}+8}}{sinx}\right)\)
1.
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{x^2-3x-4}}{1-x^2}=\lim\limits_{x\to (-1)-}\frac{\sqrt{(x+1)(x-4)}}{(1-x)(1+x)}\)
\(=\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{(x-1)\sqrt{-(x+1)}}=-\infty\) do:
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{x-1}=\frac{-\sqrt{5}}{2}<0\) và \(\lim\limits_{x\to (-1)-}\frac{1}{\sqrt{-(x+1)}}=+\infty\)
2.
\(\lim\limits_{x\to 2+}\left(\frac{1}{x-2}-\frac{x+1}{\sqrt{x+2}-2}\right)=\lim\limits_{x\to 2+}\frac{1-(x+1)(\sqrt{x+2}+2)}{x-2}=-\infty\) do:
\(\lim\limits_{x\to 2+}\frac{1}{x-2}=+\infty\) và \(\lim\limits_{x\to 2+}[1-(x+1)(\sqrt{x+2}+2)]=-11<0\)
3.
\(\lim\limits_{x\to +\infty}\frac{3x^2-5\sin 2x+7\cos ^2x}{2x^2+2}=\lim\limits_{x\to +\infty}\frac{3x^2-5\sin 2x+7(1-\sin ^2x)}{2x^2+2}\)
\(=\lim\limits_{x\to +\infty}\frac{3(x^2+1)-5\sin 2x+4-7\sin ^2x}{2x^2+2}\)
\(=\lim\limits_{x\to +\infty}\left[\frac{3}{2}-5.\frac{\sin 2x}{2x}.\frac{2x}{2x^2+2}+\frac{2}{x^2+1}-7.(\frac{\sin x}{x})^2.\frac{x^2}{2x^2+2}\right]\)
\(=\frac{3}{2}-5.0.0+0-7.0.\frac{1}{2}=\frac{3}{2}\) (nhớ rằng \(\lim\limits_{t\to \infty}\frac{\sin t}{t}=0\))
Tính các giới hạn
a) \(\lim\limits_{x\rightarrow a}\dfrac{\sin x-\sin a}{x-a}\)
b) \(\lim\limits_{x\rightarrow1}\left(1-x\right)\tan\dfrac{\pi x}{2}\)
c) \(\lim\limits_{x\rightarrow\dfrac{\pi}{3}}\dfrac{2\sin^2x+\sin x-1}{2\sin^2x-3\sin x+1}\)
d) \(\lim\limits_{x\rightarrow0}\dfrac{\tan x-\sin x}{\sin^3x}\)
Tìm \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{\left(2x+1\right)\left(3x+1\right)\left(4x+1\right)}-1}{x}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{\left(2x+1\right)\left(3x+1\right)\left(4x+1\right)}-1}{x}=\lim\limits_{x\rightarrow0}\dfrac{24x^2+26x+9}{\sqrt{\left(2x+1\right)\left(3x+1\right)\left(4x+1\right)}+1}=\dfrac{9}{1+1}=\dfrac{9}{2}\)
Ta có: \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{\left(2x+1\right)\left(3x+1\right)\left(4x+1\right)}-1}{x}\) \(=\lim\limits_{x\rightarrow0}\dfrac{\left(2x+1\right)\left(3x+1\right)\left(4x+1\right)-1}{x\left(\sqrt{\left(2x+1\right)\left(3x+1\right)\left(4x+1\right)}+1\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{x\left(24x^2+26x+9\right)}{x\left(\sqrt{\left(2x+1\right)\left(3x+1\right)\left(4x+1\right)}+1\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{24x^2+26x+9}{\sqrt{\left(2x+1\right)\left(3x+1\right)\left(4x+1\right)}+1}\)
\(=\dfrac{9}{2}\)
Tính các giới hạn sau :
a) \(\lim\limits_{x\rightarrow-3}\dfrac{x+3}{x^2+2x-3}\)
b) \(\lim\limits_{x\rightarrow0}\dfrac{\left(1+x\right)^3-1}{x}\)
c) \(\lim\limits_{x\rightarrow+\infty}\dfrac{x-1}{x^2-1}\)
d) \(\lim\limits_{x\rightarrow5}\dfrac{x-5}{\sqrt{x}-\sqrt{5}}\)
e) \(\lim\limits_{x\rightarrow+\infty}\dfrac{x-5}{\sqrt{x}+\sqrt{5}}\)
f) \(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x^2+5}-3}{x+2}\)
g) \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{x}-1}{\sqrt{x+3}-2}\)
h) \(\lim\limits_{x\rightarrow+\infty}\dfrac{1-2x+3x^3}{x^3-9}\)
i) \(\lim\limits_{x\rightarrow0}\dfrac{1}{x^2}\left(\dfrac{1}{x^2+1}-1\right)\)
j) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\left(x^2-1\right)\left(1-2x\right)^5}{x^7+x+3}\)
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)
\(I_2=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+x^2}+x\right)^n-\left(\sqrt{1+x^2}-x\right)^n}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{n\left(\sqrt{1+x^2}+x\right)^{n-1}\left(\dfrac{x}{\sqrt{1+x^2}}+1\right)-n\left(\sqrt{1+x^2}-x\right)^{n-1}\left(\dfrac{x}{\sqrt{1+x^2}}-1\right)}{1}\)
\(=\dfrac{n.1-n\left(-1\right)}{1}=2n\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-\sqrt[3]{2x+1}}{x}\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{4x+5}-3}{\sqrt[3]{5x+3}-2}\)
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt[4]{2x+3}+\sqrt[3]{2+3x}}{\sqrt{x+2}-1}\)
\(a=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-1+1-\sqrt[3]{2x+1}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x}{\sqrt[]{4x+1}+1}+\dfrac{-2x}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{4x+1}+1}+\dfrac{-2}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}\right)=...\)
\(b=\lim\limits_{x\rightarrow1}\dfrac{4\left(x-1\right)\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(x-1\right)\left(\sqrt[]{4x+5}+3\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{4\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(\sqrt[]{4x+5}+3\right)}=...\)
\(c=\lim\limits_{x\rightarrow-1}\dfrac{\left(2x+3\right)^{\dfrac{1}{4}}+\left(2+3x\right)^{\dfrac{1}{3}}}{\left(x+2\right)^{\dfrac{1}{2}}-1}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{1}{2}\left(2x+3\right)^{-\dfrac{3}{4}}+\left(2+3x\right)^{-\dfrac{2}{3}}}{\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}=3\)