\(4x^3-3x+1=\left(2x-1\right)^2\left(x+1\right)\) 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:
\(\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{\sqrt{1-3x^2}+3x-2}{4x^3-3x+1}=\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{-12\left(2x-1\right)^2}{\left(x+1\right)\left(2x-1\right)^2\left(\sqrt{1-3x^2}+2-3x\right)}\)
\(=\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{-12}{\left(x+1\right)\left(\sqrt{1-3x^2}+2-3x\right)}=-8\)
\(\Rightarrow c=-8\)
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 $x=\frac{1}{2}$
$\Leftrightarrow \sqrt{4+a}=1\Leftrightarrow a=-3$
$b=\sqrt{4+a}-4=-3$
\(\lim\limits_{x\to 0,5}\frac{\sqrt{1-3x^2}+3x-2}{4x^3-3x+1}=\lim\limits_{x\to 0,5}\frac{-3(2x-1)^2(2x+1)}{(2\sqrt{1-3x^2}+1)(\sqrt{1-3x^2}+x)(2x-1)^2(x+1)}\)
\(=\lim\limits_{x\to 0,5}\frac{-3(2x+1)}{(2\sqrt{1-3x^2}+1)(\sqrt{1-3x^2}+x)(x+1)}=-2=c\)
