Đáp án:
\[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {x + 1} - \sqrt {{x^2} + x + 1} }}{x} = 0\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {x + 1} - \sqrt {{x^2} + x + 1} }}{x}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\sqrt {x + 1} - \sqrt {{x^2} + x + 1} } \right)\left( {\sqrt {x + 1} + \sqrt {{x^2} + x + 1} } \right)}}{{x\left( {\sqrt {x + 1} + \sqrt {{x^2} + x + 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {x + 1} \right) - \left( {{x^2} + x + 1} \right)}}{{x\left( {\sqrt {x + 1} + \sqrt {{x^2} + x + 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{ - {x^2}}}{{x.\left( {\sqrt {x + 1} + \sqrt {{x^2} + x + 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{ - x}}{{\left( {\sqrt {x + 1} + \sqrt {{x^2} + x + 1} } \right)}}\\
= \dfrac{{ - 0}}{{\sqrt {0 + 1} + \sqrt {{0^2} + 0 + 1} }}\\
= 0
\end{array}\)
