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ta có (f(x)-20)/(x-2)=10
===>f(x)=10x
thay f(x)=10x vào A và thay
x=2+0,000000001 ta được giới hạn của A= -331259694,9
cái chỗ F(x) =10x đó ,đâu có là sao vậy ạ , tại có thể 10 đó là g(2)=10
\(\lim\limits_{x\rightarrow1}\dfrac{x^3-3x+2}{x^4-4x+3}=\lim\limits_{x\rightarrow1}\dfrac{\left(x+2\right)\left(x-1\right)^2}{\left(x^2+2x+3\right)\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\dfrac{x+2}{x^2+2x+3}=\dfrac{1}{2}\)
\(\lim\limits_{x\rightarrow2^-}\dfrac{x^3+x^2-4x-4}{x^2-4x+4}=\lim\limits_{x\rightarrow2^-}\dfrac{\left(x-2\right)\left(x^2+3x+2\right)}{\left(x-2\right)^2}=\lim\limits_{x\rightarrow2^-}\dfrac{x^2+3x+2}{x-2}=-\infty\)
\(\lim\limits_{x\rightarrow2}\dfrac{\left(x^2-x-2\right)^{20}}{\left(x^3-12x+16\right)^{10}}=\lim\limits_{x\rightarrow2}\dfrac{\left(x+1\right)^{20}\left(x-2\right)^{20}}{\left(x+4\right)^{10}\left(x-2\right)^{20}}=\lim\limits_{x\rightarrow2}\dfrac{\left(x+1\right)^{20}}{\left(x+4\right)^{10}}=\dfrac{3^{10}}{2^{10}}\)
\(\lim\limits_{x\rightarrow0^-}\dfrac{4x^2+5x}{x^2}=\lim\limits_{x\rightarrow0^-}\dfrac{4x+5}{x}=-\infty\)
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{x+2}-1}{\sqrt{x+5}-2}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(\sqrt{x+5}+2\right)}{\left(x+1\right)\left(\sqrt{x+2}+1\right)}=\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{x+5}+2}{\sqrt{x+2}+1}=2\)
Bài 2:
\(\lim\limits_{x\to 2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}=\lim\limits_{x\to 2}\frac{x^2-x-2}{(x+\sqrt{x+2}).\frac{4x+1-9}{\sqrt{4x+1}+3}}=\lim\limits_{x\to 2}\frac{(x-2)(x+1)(\sqrt{4x+1}+3)}{(x+\sqrt{x+2}).4(x-2)}=\lim\limits_{x\to 2}\frac{(x+1)(\sqrt{4x+1}+3)}{4(x+\sqrt{x+2})}=\frac{9}{8}\)
Bài 3:
\(\lim\limits_{x\to 0-}\frac{1-\sqrt[3]{x-1}}{x}=-\infty \)
\(\lim\limits_{x\to 0+}\frac{1-\sqrt[3]{x-1}}{x}=+\infty \)
Bài 4:
\(\lim\limits_{x\to -\infty}\frac{x^2-5x+1}{x^2-2}=\lim\limits_{x\to -\infty}\frac{1-\frac{5}{x}+\frac{1}{x^2}}{1-\frac{2}{x^2}}=1\)
Bài 5:
\(\lim\limits_{x\to +\infty}\frac{2x^2-4}{x^3+3x^2-9}=\lim\limits_{x\to +\infty}\frac{\frac{2}{x}-\frac{4}{x^3}}{1+\frac{3}{x}-\frac{9}{x^3}}=0\)
Bài 6:
\(\lim\limits_{x\to 2- }\frac{2x-1}{x-2}=\lim\limits_{x\to 2-}\frac{2(x-2)+3}{x-2}=\lim\limits_{x\to 2-}\left(2+\frac{3}{x-2}\right)=-\infty \)
Bài 7:
\(\lim\limits _{x\to 3+ }\frac{8+x-x^2}{x-3}=\lim\limits _{x\to 3+}\frac{1}{x-3}.\lim\limits _{x\to 3+}(8+x-x^2)=2(+\infty)=+\infty \)
Bài 8:
\(\lim\limits _{x\to -\infty}(8+4x-x^3)=\lim\limits _{x\to -\infty}(-x^3)=+\infty \)
Bài 9:
\(\lim\limits _{x\to -1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{x^2+3-4}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{(x-1)(x+1)}\)
\(\lim\limits _{x\to -1}\frac{\sqrt{x^2+3}+2}{(\sqrt[3]{x^2}-\sqrt[3]{x}+1)(x-1)}=\frac{-2}{3}\)
tổng chữ số hàng đơn vị và hàng chục của limx->1 \(\frac{x^{2020}-2020x+ 2019}{\left(x-1\right)^2}\)
\(\lim\limits_{x\rightarrow1}\frac{x^{2020}-2020x+2019}{\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\frac{2020x^{2019}-2020}{2\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{2019.2020.x^{2018}}{2}=\frac{2019.2020}{2}=2039190\)
Tổng chữ số hàng chục và đơn vị là 9
\(\lim\limits_{x\rightarrow-\infty}\left(4x^5-3x^2+1\right)=\lim\limits_{x\rightarrow-\infty}x^5\left(4-\frac{3}{x^3}+\frac{1}{x^5}\right)=-\infty.4=-\infty\)
\(\lim\limits_{x\rightarrow4}\frac{1-x}{\left(x-4\right)^2}=\frac{-3}{0}=-\infty\)
Câu tiếp theo đề thiếu, ko thấy yêu cầu gì hết
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 \(\lim\limits_{x\to 2+}[1-(x+1)(\sqrt{x+2}+2)]=-11<0\)
a) 
(x4 – x2 + x - 1) = x4(1 - 
) = +∞.
b) 
(-2x3 + 3x2 -5 ) = x3(-2 + 
) = +∞.
c) 
= 
= +∞.
d) \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+1}+x}{5-2x}=\lim\limits_{x\rightarrow+\infty}\dfrac{\left|x\right|\sqrt{1+\dfrac{1}{x^2}}+x}{5-2x}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x\sqrt{1+\dfrac{1}{x^2}}+x}{5-2x}\)\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x^2}}+1}{\dfrac{5}{x}-2}=-1\).
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)-\left(a+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x-a-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{x-a-1}{x^2+x+1}=\dfrac{-a}{3}\)