Cho hàm số $$f(x)=\begin{cases}
\dfrac{3-\sqrt{4-x}}{4} &\text{khi }x\neq0\\
\dfrac{1}{4} &\text{khi }x=0
\end{cases}$$Tính \(f'(0)\).
\(f'(0)=\dfrac{1}{4}\) | |
\(f'(0)=\dfrac{1}{16}\) | |
\(f'(0)=\dfrac{1}{32}\) | |
Không tồn tại |
Chọn phương án B.
\(\begin{aligned}
f'(0)&=\lim\limits_{x\to0}\dfrac{f(x)-f(0)}{x-0}\\
&=\lim\limits_{x\to0}\dfrac{\dfrac{3-\sqrt{4-x}}{4}-\dfrac{1}{4}}{x}\\
&=\lim\limits_{x\to0}\dfrac{2-\sqrt{4-x}}{4x}\\
&=\lim\limits_{x\to0}\dfrac{\left(2-\sqrt{4-x}\right)\left(2+\sqrt{4-x}\right)}{4x\left(2+\sqrt{4-x}\right)}\\
&=\lim\limits_{x\to0}\dfrac{4-(4-x)}{4x\left(2+\sqrt{4-x}\right)}\\
&=\lim\limits_{x\to0}\dfrac{x}{4x\left(2+\sqrt{4-x}\right)}\\
&=\lim\limits_{x\to0}\dfrac{1}{4\left(2+\sqrt{4-x}\right)}\\
&=\dfrac{1}{16}.
\end{aligned}\)