Đáp án:
e) \(\dfrac{3}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
d)\mathop {\lim }\limits_{x \to 0} \dfrac{{1 + {x^2} - 1}}{{x\left( {\sqrt {1 + {x^2}} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2}}}{{x\left( {\sqrt {1 + {x^2}} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 0} \dfrac{x}{{\sqrt {1 + {x^2}} + 1}} = \dfrac{0}{{1 + 1}} = 0\\
f)\mathop {\lim }\limits_{x \to 1} \dfrac{{2x + 2 - 3x - 1}}{{\left( {x - 1} \right)\left( {\sqrt {2x + 2} + \sqrt {3x + 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{ - x + 1}}{{\left( {x - 1} \right)\left( {\sqrt {2x + 2} + \sqrt {3x + 1} } \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{ - 1}}{{\sqrt {2x + 2} + \sqrt {3x + 1} }}\\
= \dfrac{{ - 1}}{{\sqrt {2 + 2} + \sqrt {3 + 1} }} = - \dfrac{1}{4}\\
e)\mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x + 2 - 4} \right)\left( {\sqrt {x + 7} + 3} \right)}}{{\left( {x + 7 - 9} \right)\left( {\sqrt {x + 2} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 2} \right)\left( {\sqrt {x + 7} + 3} \right)}}{{\left( {x - 2} \right)\left( {\sqrt {x + 2} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt {x + 7} + 3}}{{\sqrt {x + 2} + 2}} = \dfrac{{3 + 3}}{{2 + 2}} = \dfrac{3}{2}
\end{array}\)
