tìm giới hạn
\(\lim\limits_{x\rightarrow2}\frac{x^2-2x}{2\sqrt{x+2}-3x+2}\)
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NT
1
4 tháng 5 2016
\(L=\lim\limits_{x\rightarrow2}\frac{x-\sqrt{3x-2}}{x^2-4}\)
\(=\lim\limits_{x\rightarrow2}\frac{x^2-3x+2}{\left(x-4\right)\left(x+\sqrt{3x-2}\right)}=\lim\limits_{x\rightarrow2}\frac{\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x+2\right)\left(x+\sqrt{3x-2}\right)}\)
\(=\lim\limits_{x\rightarrow2}\frac{x-1}{\left(x+2\right)\left(x+\sqrt{3x-2}\right)}=\frac{1}{16}\)
HT
24 tháng 1 2021
a/ L'Hospital:
\(=\lim\limits_{x\rightarrow2}\dfrac{x-\left(x+2\right)^{\dfrac{1}{2}}}{\left(4x+1\right)^{\dfrac{1}{2}}-3}=\lim\limits_{x\rightarrow2}\dfrac{1-\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}{\dfrac{1}{2}\left(4x+1\right)^{-\dfrac{1}{2}}.4}=\dfrac{1-\dfrac{1}{2}.4^{-\dfrac{1}{2}}}{2.9^{-\dfrac{1}{2}}}=\dfrac{9}{8}\)
b/ L'Hospital:\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7\right)^{\dfrac{1}{2}}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{2}\left(2x+7\right)^{-\dfrac{1}{2}}.2+1}{3x^2-8x}=\dfrac{9^{-\dfrac{1}{2}}+1}{3-8}=-\dfrac{4}{15}\)
6 tháng 12 2023
a: \(\lim\limits_{x\rightarrow2}\dfrac{1-\sqrt{x^2+3}}{-x^2+3x-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\sqrt{x^2+3}-1}{x^2-3x+2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\sqrt{2^2+3}-1}{2^2-3\cdot2+2}\)
\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}\sqrt{2^2+3}-1=\sqrt{7}-1>0\\\lim\limits_{x\rightarrow2}2^2-3\cdot2+2=0\end{matrix}\right.\)
b: \(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{4x-1}+3}{x^2-4}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{4x-1-9}{\sqrt{4x-1}-3}\cdot\dfrac{1}{x^2-4}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{4x-10}{\sqrt{4x-1}-3}\cdot\dfrac{1}{\left(x-2\right)\left(x+2\right)}\)
\(=+\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}\dfrac{4x-10}{\sqrt{4x-1}-3}=\dfrac{4\cdot2-10}{\sqrt{4\cdot2-1}-3}=\dfrac{-2}{\sqrt{7}-3}>0\\\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x-2\right)\cdot\left(x+2\right)}=\dfrac{1}{\left(2+2\right)\cdot\left(2-2\right)}=+\infty\end{matrix}\right.\)
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 \(3-x>0;\forall x< 3\)
\(\Rightarrow\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}=-\infty\)
10 tháng 11 2023
a: \(\lim\limits_{x\rightarrow1}\dfrac{x^2-1}{\sqrt{3x+1}-2}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)}{\dfrac{3x+1-4}{\sqrt{3x+1}+2}}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)\cdot\left(\sqrt{3x+1}+2\right)}{3\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x+1\right)\left(\sqrt{3x+1}+2\right)}{3}\)
\(=\dfrac{\left(1+1\right)\left(\sqrt{3+1}+2\right)}{2}=\dfrac{2\cdot4}{3}=\dfrac{8}{3}\)
b: \(\lim\limits_{x\rightarrow2}\dfrac{x^2-2x}{\sqrt{x+2}-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)}{\dfrac{x+2-4}{\sqrt{x+2}+2}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)\cdot\left(\sqrt{x+2}+2\right)}{x-2}\)
\(=\lim\limits_{x\rightarrow2}x\left(\sqrt{x+2}+2\right)\)
\(=2\cdot\left(\sqrt{2+2}+2\right)\)
\(=2\cdot4=8\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 2 2020
Tất cả đều ko phải dạng vô định, bạn cứ thay số vào tính thôi:
\(a=\frac{sin\left(\frac{\pi}{4}\right)}{\frac{\pi}{2}}=\frac{\sqrt{2}}{\pi}\)
\(b=\frac{\sqrt[3]{3.4-4}-\sqrt{6-2}}{3}=\frac{0}{3}=0\)
\(c=0.sin\frac{1}{2}=0\)
10 tháng 11 2023
a: \(\lim\limits_{x\rightarrow4}\dfrac{\sqrt{2x+8}-4}{x-4}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{2x+8-16}{\sqrt{2x+8}+4}\cdot\dfrac{1}{x-4}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{2\left(x-4\right)}{\sqrt{2x+8}+4}\cdot\dfrac{1}{x-4}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{2}{\sqrt{2x+8}+4}=\dfrac{2}{\sqrt{2\cdot4+8}+4}\)
\(=\dfrac{2}{\sqrt{8+8}+4}=\dfrac{2}{4+4}=\dfrac{2}{8}=\dfrac{1}{4}\)
b: \(\lim\limits_{x\rightarrow2}\dfrac{x^2-4}{\sqrt{4x+1}-3}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{\dfrac{4x+1-9}{\sqrt{4x+1}+3}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{4\left(x-2\right)}\cdot\left(\sqrt{4x+1}+3\right)\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x+2\right)\left(\sqrt{4x+1}+3\right)}{4}\)
\(=\dfrac{\left(2+2\right)\left(\sqrt{4\cdot2+1}+3\right)}{4}=\sqrt{9}+3=6\)
c: \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-\sqrt{x+2}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\dfrac{4-x-2}{2+\sqrt{x+2}}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-x}\cdot\left(\sqrt{x+2}+2\right)\)
\(=\lim\limits_{x\rightarrow2}\left(-\sqrt{x+2}-2\right)\)
\(=-\sqrt{2+2}-2=-2-2=-4\)
6 tháng 12 2023
\(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt[3]{2x+12}+x}{x^2+2x}\)
\(=\lim\limits_{x\rightarrow-2}\left(\dfrac{2x+12+x^3}{\sqrt[3]{\left(2x+12\right)^2}-x\cdot\sqrt[3]{2x+12}+x^2}\cdot\dfrac{1}{x^2+2x}\right)\)
\(=\lim\limits_{x\rightarrow-2}\left(\dfrac{x^3+2x^2-2x^2-4x+6x+12}{\left(\sqrt[3]{\left(2x+12\right)^2}+x\cdot\sqrt[3]{2x+12}+x^2\right)\cdot x\cdot\left(x+2\right)}\right)\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{\left(x+2\right)\left(x^2-2x+6\right)}{\left(\sqrt[3]{\left(2x+12\right)^2}-x\cdot\sqrt[3]{2x+12}+x^2\right)\cdot x\cdot\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{x^2-2x+6}{x\cdot\left(\sqrt[3]{\left(2x+12\right)^2}-x\cdot\sqrt[3]{2x+12}+x^2\right)}\)
\(=\dfrac{\left(-2\right)^2-2\cdot\left(-2\right)+6}{\sqrt[3]{\left(-2\cdot2+12\right)^2}-\left(-2\right)\cdot\sqrt[3]{2\cdot\left(-2\right)+12}+\left(-2\right)^2}\)
\(=\dfrac{4+4+6}{\sqrt[3]{64}+2\cdot\sqrt[3]{8}+4}\)
\(=\dfrac{14}{8+2\cdot2+4}=\dfrac{14}{12+4}=\dfrac{14}{16}=\dfrac{7}{8}\)
b: \(\lim\limits_{x\rightarrow2}\dfrac{x-\sqrt{x+2}}{x^3-8}\)
\(=\lim\limits_{x\rightarrow2}\left(\dfrac{x^2-x-2}{x+\sqrt{x+2}}\cdot\dfrac{1}{x^3-8}\right)\)
\(=\lim\limits_{x\rightarrow2}\left(\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x+\sqrt{x+2}\right)\cdot\left(x-2\right)\left(x^2+2x+4\right)}\right)\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x+1}{\left(x+\sqrt{x+2}\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{2+1}{\left(2+\sqrt{2+2}\right)\left(2^2+2\cdot2+4\right)}\)
\(=\dfrac{3}{\left(2+2\right)\left(4+4+4\right)}=\dfrac{3}{12\cdot4}=\dfrac{1}{4\cdot4}=\dfrac{1}{16}\)
23 tháng 7 2023
a: \(=\lim\limits_{x\rightarrow+\infty}\dfrac{4+\dfrac{3}{x}}{2}=\dfrac{4}{2}=2\)
b: \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{2}{x}}{3+\dfrac{1}{x}}=0\)
c: \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x^2}}}{1+\dfrac{1}{x}}=1\)
Bạn tự hiểu là giới hạn khi x tới 2:
\(=\frac{x\left(x-2\right)\left[2\sqrt{x+2}+3x-2\right]}{4\left(x+2\right)-\left(3x-2\right)^2}=\frac{x\left(x-2\right)\left[2\sqrt{x+2}+3x-2\right]}{-9x^2+16x+4}=\frac{x\left(x-2\right)\left[2\sqrt{x+2}+3x-2\right]}{\left(x-2\right)\left(-9x-2\right)}\)
\(=\frac{x\left[2\sqrt{x+2}+3x-2\right]}{-9x-x}=\frac{2\left[2\sqrt{4}+6-2\right]}{-18-2}=...\)