a) Đúng

b) Đúng

c) Đúng

d) Sai

a) \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {4 + x} - 2}}{{4x}} = \mathop {\lim }\limits_{x \to 0} \frac{{(\sqrt {4 + x} - 2)(\sqrt {4 + x} + 2)}}{{4x(\sqrt {4 + x} + 2)}} = \mathop {\lim }\limits_{x \to 0} \frac{{4 + x - 4}}{{4x(\sqrt {4 + x} + 2)}}\)

\( = \mathop {\lim }\limits_{x \to 0} \frac{1}{{4(\sqrt {4 + x} + 2)}} = \frac{1}{{4(\sqrt 4 + 2)}} = \frac{1}{{16}}{\rm{. }}\)

b)

\(\begin{array}{*{20}{l}}{\mathop {\lim }\limits_{x \to 2} \frac{{4 - {x^2}}}{{\sqrt {x + 7} - 3}}}&{ = \mathop {\lim }\limits_{x \to 2} \frac{{(2 - x)(2 + x)(\sqrt {x + 7} + 3)}}{{(\sqrt {x + 7} - 3)(\sqrt {x + 7} + 3)}} = \mathop {\lim }\limits_{x \to 2} \frac{{(2 - x)(2 + x)(\sqrt {x + 7} + 3)}}{{x + 7 - 9}}}\\{}&{ = \mathop {\lim }\limits_{x \to 2} [ - (2 + x)(\sqrt {x + 7} + 3)] = - 4.6 = - 24}\end{array}\)

c)

\({\rm{ }}\begin{array}{*{20}{l}}{\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {2x + 5} - 3}}{{\sqrt {x + 2} - 2}}}&{ = \mathop {\lim }\limits_{x \to 2} \frac{{(\sqrt {2x + 5} - 3)(\sqrt {2x + 5} + 3)(\sqrt {x + 2} + 2)}}{{(\sqrt {x + 2} - 2)(\sqrt {x + 2} + 2)(\sqrt {2x + 5} + 3)}}}\\{}&{ = \mathop {\lim }\limits_{x \to 2} \frac{{(2x + 5 - 9)(\sqrt {x + 2} + 2)}}{{(x + 2 - 4)(\sqrt {2x + 5} + 3)}} = \mathop {\lim }\limits_{x \to 2} \frac{{2(\sqrt {x + 2} + 2)}}{{\sqrt {2x + 5} + 3}} = \frac{4}{3}}\end{array}\)

d) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt[3]{{x + 7}} - 2}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{(\sqrt[3]{{x + 7}} - 2)\left( {\sqrt[3]{{{{(x + 7)}^2}}} + 2\sqrt[3]{{x + 7}} + 4} \right)}}{{(x - 1)\left( {\sqrt[3]{{{{(x + 7)}^2}}} + 2\sqrt[3]{{x + 7}} + 4} \right)}}\)

\( = \mathop {\lim }\limits_{x \to 1} \frac{{x + 7 - {2^3}}}{{(x - 1)\left( {\sqrt[3]{{{{(x + 7)}^2}}} + 2\sqrt[3]{{x + 7}} + 4} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt[3]{{{{(x + 7)}^2}}} + 2\sqrt[3]{{x + 7}} + 4}} = \frac{1}{{12}}.\)

Cho lim x → 2 + ( x − 2 ) √ x /(x^ 2 − 4) . Tính giới hạn đó.