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23 tháng 12 2023
\(\lim\limits_{x\rightarrow2}\dfrac{\left(3x-5\right)}{\left(x-2\right)^2}=+\infty\)
vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2}3x-5=3\cdot2-5=1>0\\\left(x-2\right)^2>0\\\lim\limits_{x\rightarrow2}\left(x-2\right)^2=\left(2-2\right)^2=0\end{matrix}\right.\)
TL
7 tháng 11 2023
Em là tám lại ạ
Em là duy khôi ạ
Em là văn tam ạ
Em là mạnh Tuấn ạ
8 tháng 11 2023
a: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x-2}-\dfrac{12}{x^3-8}\right)\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x+4-12}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x-8}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x+4}{x^2+2x+4}\)
\(=\dfrac{2+4}{2^2+2\cdot2+4}=\dfrac{6}{4+4+4}=\dfrac{6}{12}=\dfrac{1}{2}\)
b: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\)
\(=\lim\limits_{x\rightarrow2}\left(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}\right)\)
\(=\lim\limits_{x\rightarrow2}\left(\dfrac{x-3+x-1}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\right)\)
\(=\lim\limits_{x\rightarrow2}\dfrac{2x-4}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{2}{\left(x-3\right)\left(x-1\right)}=\dfrac{2}{\left(2-3\right)\left(2-1\right)}=-2\)
d: \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-\sqrt[3]{x^3-1}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-x+x-\sqrt[3]{x^3-1}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}+\dfrac{x^3-x^3+1}{\sqrt[3]{x^2}+x\cdot\sqrt[3]{x^3-1}+\sqrt[3]{\left(x^3-1\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{1}{\sqrt{x^2+1}+x}+\dfrac{1}{\sqrt[3]{x^2}+x\cdot\sqrt[3]{x^3-1}+\sqrt[3]{\left(x^3-1\right)^2}}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\dfrac{1}{x}}{\sqrt{1+\dfrac{1}{x^2}}+1}+\dfrac{\dfrac{1}{x^2}}{\sqrt[3]{\dfrac{1}{x^4}}+\sqrt[3]{1-\dfrac{1}{x^3}}+\sqrt[3]{\left(1-\dfrac{1}{x^3}\right)^2}}\right)\)
=0
c: \(\lim\limits_{x\rightarrow+\infty}\left[x\cdot\left(\sqrt{x^2+1}-x\right)\right]\)
\(=\lim\limits_{x\rightarrow+\infty}\left[x\cdot\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}\right]\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x}{\sqrt{x^2+1}+x}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}+1}=\dfrac{1}{1+1}=\dfrac{1}{2}\)
e: \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{x^2+1-1}{\sqrt{x^2+1}+1}:\dfrac{x^2+16-16}{\sqrt{x^2+16}+4}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+16}+4}{\sqrt{x^2+1}+1}=\dfrac{4+4}{1+1}=\dfrac{8}{2}=4\)
NM
1
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 2 2020
\(=\lim\limits_{x\rightarrow2^-}\frac{-\left(x+2\right)\sqrt{\left(2-x\right)^2}}{\sqrt{\left(x^2+1\right)\left(2-x\right)}}=\lim\limits_{x\rightarrow2^-}\frac{-\left(x+2\right)\sqrt{2-x}}{\sqrt{x^2+1}}=\frac{0}{\sqrt{5}}=0\)
C
1
24 tháng 12 2023
\(\lim\limits_{x\rightarrow2^+}\left(\dfrac{1}{x^2-4}-\dfrac{1}{x-2}\right)\)
\(=\lim\limits_{x\rightarrow2^+}\left(\dfrac{1}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right)\)
\(=\lim\limits_{x\rightarrow2^+}\left(\dfrac{1-x-2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\lim\limits_{x\rightarrow2^+}\left(\dfrac{-x-1}{x^2-4}\right)\)
\(=-\infty\) vì \(\left\{{}\begin{matrix}x^2-4>0\\\lim\limits_{x\rightarrow2^+}x^2-4=2^2-4=0\\\lim\limits_{x\rightarrow2^+}-x-1=-2-1=-3< 0\end{matrix}\right.\)
NM
1
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 3 2023
Giới hạn đã cho hữu hạn nên \(x^2+2ax-b=0\) có nghiệm \(x=2\)
\(\Rightarrow4+4a-b=0\Rightarrow b=4a+4\)
\(\Rightarrow\lim\limits_{x\rightarrow2}\dfrac{x^2+2ax-4a-4}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2a+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x+2a+2}{x+2}=\dfrac{2a+4}{4}=4\)
\(\Rightarrow a=6\Rightarrow b=28\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2022
\(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)+1}{x-2}\) hữu hạn \(\Rightarrow f\left(x\right)+1=0\) có nghiệm \(x=2\Rightarrow f\left(2\right)=-1\)
\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{f\left(x\right)+2x+1}-x}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{1}{\sqrt{f\left(x\right)+2x+1}+x}.\dfrac{\left(\sqrt{f\left(x\right)+2x+1}-x\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}.\dfrac{f\left(x\right)+1-x\left(x-2\right)}{x-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}.\left(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)+1}{x-2}-\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)}{x-2}\right)\)
\(=\dfrac{1}{4\left(\sqrt{4}+2\right)}.\left(a-2\right)=\dfrac{a-2}{16}\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 1 2021
\(=\lim\limits_{x\rightarrow2^-}\dfrac{-\left(2-x\right)\left(2+x\right)}{\sqrt{\left(x^4+1\right)\left(2-x\right)}}=\lim\limits_{x\rightarrow2^-}\dfrac{-\left(2+x\right)\sqrt{2-x}}{\sqrt{x^4+1}}=\dfrac{0}{3}=0\)
\(\lim\limits_{x\rightarrow2^-}\left(\dfrac{1}{x-2}-\dfrac{1}{x^2-4}\right)\)
\(=\lim\limits_{x\rightarrow2^-}\dfrac{x+2-1}{\left(x-2\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow2^-}\dfrac{x+1}{\left(x-2\right)\left(x+2\right)}\)
\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow2^-}\dfrac{x+1}{x+2}=\dfrac{2+1}{2+2}=\dfrac{3}{4}>0\\x-2< 0\end{matrix}\right.\)