Đáp án:
h) \(\dfrac{1}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\mathop {\lim }\limits_{x \to - 3} \dfrac{{x + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \mathop {\lim }\limits_{x \to - 3} \dfrac{1}{{x - 3}}\\
= \dfrac{1}{{ - 3 - 3}} = - \dfrac{1}{6}\\
e)\mathop {\lim }\limits_{x \to 2} \dfrac{{x + 7 - 9}}{{\left( {x - 2} \right)\left( {\sqrt {x + 7} + 3} \right)}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{x - 2}}{{\left( {x - 2} \right)\left( {\sqrt {x + 7} + 3} \right)}}\\
= \mathop {\lim }\limits_{x \to 2} \dfrac{1}{{\sqrt {x + 7} + 3}} = \dfrac{1}{{\sqrt {2 + 7} + 3}} = \dfrac{1}{6}\\
b)\mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 2} \right)\left( {x + 3} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{x + 3}}{{x + 2}}\\
= \dfrac{{2 + 3}}{{2 + 2}} = \dfrac{5}{4}\\
f)\mathop {\lim }\limits_{x \to 1} \dfrac{{x + 3 - 4}}{{\left( {x - 1} \right)\left( {\sqrt {x + 3} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{x - 1}}{{\left( {x - 1} \right)\left( {\sqrt {x + 3} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{1}{{\sqrt {x + 3} + 2}}\\
= \dfrac{1}{{\sqrt {1 + 3} + 2}} = \dfrac{1}{4}\\
c)\mathop {\lim }\limits_{x \to 4} \dfrac{{\left( {x - 4} \right)\left( {x + 4} \right)}}{{\left( {x - 4} \right)\left( {x + 5} \right)}} = \mathop {\lim }\limits_{x \to 4} \dfrac{{x + 4}}{{x + 5}} = \dfrac{8}{9}\\
d)\mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{x + 2}}{{x - 1}} = 4\\
h)\mathop {\lim }\limits_{x \to - 3} \dfrac{{x + 7 - 4}}{{\left( {x + 3} \right)\left( {\sqrt {x + 7} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to - 3} \dfrac{1}{{\sqrt {x + 7} + 2}} = \dfrac{1}{{\sqrt { - 3 + 7} + 2}} = \dfrac{1}{4}
\end{array}\)
