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