Question

[Julian Edward]

cho \(\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{\sqrt{1+ax^2}-bx-2}{4x^3-3x+1}=c\)  (a,b,c thuoc R). tìm a, b, c?

Answer 1

[Nguyễn Việt Lâm]

\(4x^3-3x+1=\left(2x-1\right)^2\left(x+1\right)\) có nghiệm kép \(x=\dfrac{1}{2}\)

\(\Rightarrow\sqrt{1+ax^2}-bx-2=0\) có nhiều hơn 1 nghiệm \(x=\dfrac{1}{2}\)

\(\Rightarrow\sqrt{1+\dfrac{a}{4}}=\dfrac{b}{2}+2\Rightarrow\sqrt{a+4}=b+4\) (\(b\ge-4\))

\(\Rightarrow a=b^2+8b+12\)

\(\Rightarrow\sqrt{1+\left(b^2+8b+12\right)x^2}=bx+2\)

\(\Rightarrow1+\left(b^2+8b+12\right)x^2=b^2x^2+4bx+4\)

\(\Rightarrow\left(8b+12\right)x^2-4bx-3=0\)

\(\Rightarrow\left(2x-1\right)\left[\left(4b+6\right)x+3\right]=0\)

\(\Rightarrow\left(4b+6\right)x+3=0\) có nghiệm \(x=\dfrac{1}{2}\)

\(\Rightarrow2b+3+3=0\Rightarrow b=-3\) \(\Rightarrow a=-3\)

Khi đó:

\(\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{\sqrt{1-3x^2}+3x-2}{4x^3-3x+1}=\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{-12\left(2x-1\right)^2}{\left(x+1\right)\left(2x-1\right)^2\left(\sqrt{1-3x^2}+2-3x\right)}\)

\(=\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{-12}{\left(x+1\right)\left(\sqrt{1-3x^2}+2-3x\right)}=-8\)

\(\Rightarrow c=-8\)

Answer 2

[Akai Haruma]

Lời giải:

\(\lim\limits_{x\to 0,5}\frac{\sqrt{1+ax^2}-bx-2}{4x^3-3x+1}=\lim\limits_{x\to 0,5}\frac{\sqrt{1+ax^2}-bx-2}{(x+1)(2x-1)^2}\)

Để giới hạn hàm đã cho hữu hạn thì $f(x)=\sqrt{1+ax^2}-bx-2$ có nhân tử là $(2x-1)^2$

$f(x)$ có nhân tử $2x-1 \Leftrightarrow f(\frac{1}{2})=0\Leftrightarrow b=\sqrt{4+a}-4$

Khi đó:

$\sqrt{1+ax^2}-bx-2=(2x-1)(2-\frac{2x+1}{\sqrt{1+ax^2}+x\sqrt{4+a}})$

Giờ ta cần xác định $a,b$ để $2-\frac{2x+1}{\sqrt{1+ax^2}+x\sqrt{4+a}}=0$ với $x=\frac{1}{2}$

$\Leftrightarrow \sqrt{4+a}=1\Leftrightarrow a=-3$

$b=\sqrt{4+a}-4=-3$

\(\lim\limits_{x\to 0,5}\frac{\sqrt{1-3x^2}+3x-2}{4x^3-3x+1}=\lim\limits_{x\to 0,5}\frac{-3(2x-1)^2(2x+1)}{(2\sqrt{1-3x^2}+1)(\sqrt{1-3x^2}+x)(2x-1)^2(x+1)}\)

\(=\lim\limits_{x\to 0,5}\frac{-3(2x+1)}{(2\sqrt{1-3x^2}+1)(\sqrt{1-3x^2}+x)(x+1)}=-2=c\)