Câu hỏi của Dang Tung

Question

Nguyễn Việt Lâm · Accepted Answer

Do $\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-1}{x^2-3x+2}=1$ hữu hạn nên $f\left(x\right)-1=0$ có nghiệm $x=1$

$\Rightarrow f\left(1\right)=1$

$\Rightarrow\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-\sqrt{2-f\left(x\right)}}{1-x^2}=\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-1+1-\sqrt{2-f\left(x\right)}}{1-x^2}$

$=\lim\limits_{x\rightarrow1}\left(\dfrac{f\left(x\right)-1}{x^2-3x+2}.\dfrac{2-x}{x+1}+\dfrac{f\left(x\right)-1}{x^2-3x+2}.\dfrac{2-x}{\left(x+1\right)\left(1+\sqrt{2-f\left(x\right)}\right)}\right)$

$=1.\dfrac{2-1}{1+1}+1.\dfrac{2-1}{\left(1+1\right).\left(1+\sqrt{2-f\left(1\right)}\right)}=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4}$

Cách 2: $\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-1}{x^2-3x+2}=1\Rightarrow$ chọn $f\left(x\right)=2-x$

Khi đó:

$\lim\limits_{x\rightarrow1}\dfrac{2-x-\sqrt{x}}{1-x^2}=\dfrac{3}{4}$

Câu trả lời:

Do $\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-1}{x^2-3x+2}=1$ hữu hạn nên $f\left(x\right)-1=0$ có nghiệm $x=1$

$\Rightarrow f\left(1\right)=1$

$\Rightarrow\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-\sqrt{2-f\left(x\right)}}{1-x^2}=\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-1+1-\sqrt{2-f\left(x\right)}}{1-x^2}$

$=\lim\limits_{x\rightarrow1}\left(\dfrac{f\left(x\right)-1}{x^2-3x+2}.\dfrac{2-x}{x+1}+\dfrac{f\left(x\right)-1}{x^2-3x+2}.\dfrac{2-x}{\left(x+1\right)\left(1+\sqrt{2-f\left(x\right)}\right)}\right)$

$=1.\dfrac{2-1}{1+1}+1.\dfrac{2-1}{\left(1+1\right).\left(1+\sqrt{2-f\left(1\right)}\right)}=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4}$

Cách 2: $\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-1}{x^2-3x+2}=1\Rightarrow$ chọn $f\left(x\right)=2-x$

Khi đó:

$\lim\limits_{x\rightarrow1}\dfrac{2-x-\sqrt{x}}{1-x^2}=\dfrac{3}{4}$

x	– 100 000	– 10 000	– 1 000	– 100	– 10
y = f(x)	?	?	?	–0,01	–0,1

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