Câu 2.86 trang 84 sách bài tập Giải tích 12 Nâng cao

Tìm các giới hạn sau:

a) \(\mathop {\lim }\limits_{x \to 0} {{{e^{5x + 3}} - {e^3}} \over {2x}}\)                                   b) \(\mathop {\lim }\limits_{x \to 0} {{{e^x} - 1} \over {\sqrt {x + 1}  - 1}}\)

c) \(\mathop {\lim }\limits_{x \to 0} {{\ln \left( {1 + {x^3}} \right)} \over {2x}}\)                                   d) \(\mathop {\lim }\limits_{x \to 0} {{\ln \left( {1 + 2x} \right)} \over {\tan x}}\)

Giải

a) \(\mathop {\lim }\limits_{x \to 0} {{{e^3}\left( {{e^{5x}} - 1} \right)} \over {5x}}.{5 \over 2} = {5 \over 2}{e^3}\mathop {\lim }\limits_{x \to 0} {{{e^{5x}} - 1} \over {5x}} = {5 \over 2}{e^3}\)

b) \(\mathop {\lim }\limits_{x \to 0} {{{e^x} - 1} \over {\sqrt {x + 1}  - 1}}\)

\(=\mathop {\lim }\limits_{x \to 0} {{{e^x} - 1} \over {\sqrt {x + 1}  - 1}} = \mathop {\lim }\limits_{x \to 0} {{({e^x} - 1)(\sqrt {x + 1}  + 1)} \over x}\)

\( = \mathop {\lim }\limits_{x \to 0} {{{e^x} - 1} \over x}.(\sqrt {x + 1}  + 1) = 2\)

c) \(\mathop {\lim }\limits_{x \to 0} {{\ln \left( {1 + {x^3}} \right)} \over {2x}}\)

\(=\mathop {\lim }\limits_{x \to 0} {{\ln \left( {{x^3} + 1} \right)} \over {{x^3}}} \cdot {1 \over 2}{x^2} = 1.0 = 0\)

d)  \(\mathop {\lim }\limits_{x \to 0} {{\ln \left( {1 + 2x} \right)} \over {\tan x}} = \mathop {\lim }\limits_{x \to 0} {{{{\ln \left( {1 + 2x} \right)} \over {2x}}} \over {{{\tan x} \over x}}}.2 = {1 \over 1}.2 = 2\)

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