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NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 3 2022
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-\left(x+1\right)}{2x^2-x}=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{x^2+1}-\left(x+1\right)\right)\left(\sqrt{x^2+1}+x+1\right)}{x\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{-2x}{x\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{-2}{\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)
\(=\dfrac{-2}{\left(0-1\right)\left(\sqrt{1}+1\right)}=1\)
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 \(3-x>0;\forall x< 3\)
\(\Rightarrow\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}=-\infty\)
AN
1
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 6 2020
\(=\lim\limits_{x\rightarrow0}\frac{-2sin\frac{4037x}{2}.sin\frac{x}{2}}{x}=\lim\limits_{x\rightarrow0}\left(-sin\frac{4037x}{2}.\frac{sin\frac{x}{2}}{\frac{x}{2}}\right)=0.1=0\)
Mẫu là \(x^2\) thì có vẻ đẹp hơn :)
NN
1
AH
Akai Haruma
Giáo viên
10 tháng 4 2018
Lời giải:
Sử dụng liên hợp:
\(\lim_{x\to 0}\frac{\sqrt{x+1}+\sqrt{x+4}-3}{x}=\lim_{x\to 0}\frac{(\sqrt{x+1}-1)+(\sqrt{x+4}-2)}{x}\)
\(=\lim_{x\to 0}\frac{\frac{x}{\sqrt{x+1}+1}+\frac{x}{\sqrt{x+4}+2}}{x}=\lim_{x\to 0}\frac{1}{\sqrt{x+1}+1}+\frac{1}{\sqrt{x+4}+2}\)
\(=\frac{1}{\sqrt{0+1}+1}+\frac{1}{\sqrt{0+4}+2}=\frac{3}{4}\)
HT
6 tháng 2 2021
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
\(N=\lim\limits_{x\rightarrow0}\dfrac{\left(1+ax\right)^{\dfrac{1}{m}}-\left(1+bx\right)^{\dfrac{1}{n}}}{\left(1+x\right)^{\dfrac{1}{2}}-1}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{m}.\left(1+ax\right)^{\dfrac{1}{m}-1}.a-\dfrac{1}{n}\left(1+bx\right)^{\dfrac{1}{n}-1}.b}{\dfrac{1}{2}\left(1+x\right)^{-\dfrac{1}{2}}}=\dfrac{\dfrac{a}{m}-\dfrac{b}{n}}{\dfrac{1}{2}}\)
\(V=\lim\limits_{x\rightarrow0}\dfrac{\left(1+mx\right)^n-\left(1+nx\right)^m}{\left(1+2x\right)^{\dfrac{1}{2}}-\left(1+3x\right)^{\dfrac{1}{3}}}=\lim\limits_{x\rightarrow0}\dfrac{n\left(1+mx\right)^{n-1}.m-m\left(1+nx\right)^{m-1}.n}{\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{1}{2}}.2-\dfrac{1}{3}\left(1+3x\right)^{-\dfrac{2}{3}}.3}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{n\left(n-1\right)\left(1+mx\right)^{n-2}.m-m\left(m-1\right)\left(1+nx\right)^{m-2}.n}{-\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{3}{2}}.2+\dfrac{2}{9}.3.3\left(1+3x\right)^{-\dfrac{5}{3}}}=....\left(thay-x-vo-la-duoc\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2020
\(a=\frac{0-1}{0-1}=1\)
\(b=\lim\limits_{x\rightarrow0}\frac{\frac{x^2}{\sqrt[3]{\left(1+x^2\right)^2}+\sqrt[3]{1+x^2}+1}}{x^2}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt[3]{\left(1+x^2\right)^2}+\sqrt[3]{1+x^2}+1}=\frac{1}{3}\)
\(c=\lim\limits_{x\rightarrow2}\frac{\sqrt{x+2}-2+\sqrt{x+7}-3}{x-2}=\lim\limits_{x\rightarrow2}\frac{\frac{x-2}{\sqrt{x+2}+2}+\frac{x-2}{\sqrt{x+7}+3}}{x-2}=\lim\limits_{x\rightarrow2}\left(\frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{x+7}+3}\right)\)
\(=\frac{1}{\sqrt{4}+2}+\frac{1}{\sqrt{9}+3}=\frac{5}{12}\)
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 3 2020
\(a=\lim\limits_{x\rightarrow0}\frac{x^2}{x\left(\sqrt{1+x^2}+1\right)}=\lim\limits_{x\rightarrow0}\frac{x}{\sqrt{1+x^2}+1}=\frac{0}{2}=0\)
\(b=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+7}-2+2-\sqrt{5-x^2}}{x-1}=\lim\limits_{x\rightarrow1}\frac{\frac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{\left(x-1\right)\left(x+1\right)}{2+\sqrt{5-x^2}}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\left(\frac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{x+1}{2+\sqrt{5-x^2}}\right)=\frac{1}{12}+\frac{1}{2}=\frac{7}{12}\)
\(c=\lim\limits_{x\rightarrow0}\frac{2x}{x\left(\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{\left(1+x\right)\left(1-x\right)}+\sqrt[3]{\left(1-x\right)^2}\right)}=\lim\limits_{x\rightarrow0}\frac{2}{\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{\left(1+x\right)\left(1-x\right)}+\sqrt[3]{\left(1-x\right)^2}}=\frac{2}{3}\)
\(d=\frac{\sqrt[3]{6}}{0}=+\infty\)
Ko sử dụng L'Hopital:
\(\lim\limits_{x\rightarrow0}\dfrac{1-cosx+cosx\left(1-cos2x\right)}{x^2}=\lim\limits_{x\rightarrow0}\dfrac{2sin^2\dfrac{x}{2}+2cosx.sin^2x}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{1}{2}\left(\dfrac{sin\dfrac{x}{2}}{\dfrac{x}{2}}\right)^2+\lim\limits_{x\rightarrow0}2cosx\left(\dfrac{sinx}{x}\right)^2=\dfrac{1}{2}.1^2+2.1.1^2=\dfrac{5}{2}\)