Đáp án:
$f'\left( 0 \right) = \dfrac{1}{2}$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right) - f\left( x \right)}}{{x - 0}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{\sqrt {{x^3} + {x^2} + 1} - 1}}{x} - 0}}{{x - 0}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^3} + {x^2} + 1} - 1}}{{{x^2}}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^3} + {x^2}}}{{{x^2}\left( {\sqrt {{x^3} + {x^2} + 1} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{x + 1}}{{\sqrt {{x^3} + {x^2} + 1} + 1}}\\
= \dfrac{{0 + 1}}{{\sqrt {{0^3} + {0^2} + 1} + 1}}\\
= \dfrac{1}{2}\\
\Rightarrow f'\left( 0 \right) = \dfrac{1}{2}
\end{array}$
Vậy $f'\left( 0 \right) = \dfrac{1}{2}$
