Hàm số \(f(x)=\begin{cases}\dfrac{\sqrt{1-3x+x^2}-\sqrt{1+x}}{x} &\text{khi }x\neq0\\
m &\text{khi }x=0\end{cases}\) liên tục tại \(x_0=0\) khi
 | \(m=4\) |
 | \(m=-1\) |
 | \(m=3\) |
 | \(m=-2\) |
Chọn phương án D.
\(\begin{aligned}
\blacksquare\,\lim\limits_{x\to0}f(x)&=\lim\limits_{x\to0}\dfrac{\sqrt{1-3x+x^2}-\sqrt{1+x}}{x}\\
&=\lim\limits_{x\to0}\dfrac{\left(\sqrt{1-3x+x^2}-\sqrt{1+x}\right)\left(\sqrt{1-3x+x^2}+\sqrt{1+x}\right)}{x\left(\sqrt{1-3x+x^2}+\sqrt{1+x}\right)}\\
&=\lim\limits_{x\to0}\dfrac{\left(1-3x+x^2\right)-\left(1+x\right)}{x\left(\sqrt{1-3x+x^2}+\sqrt{1+x}\right)}\\
&=\lim\limits_{x\to0}\dfrac{x^2-4x}{x\left(\sqrt{1-3x+x^2}+\sqrt{1+x}\right)}\\
&=\lim\limits_{x\to0}\dfrac{x-4}{\sqrt{1-3x+x^2}+\sqrt{1+x}}\\
&=\dfrac{0-4}{\sqrt{1-3\cdot0+0^2}+\sqrt{1+0}}\\
&=-2.
\end{aligned}\)
\(\blacksquare\,f(0)=m\).
Để \(f(x)\) liên tục tại \(x=0\) thì $$\lim\limits_{x\to0}f(x)=f(0)\Leftrightarrow m=-2.$$