Tìm các giới hạn sau:
a) lim 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}}
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=\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