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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) \(\mathop {\lim }\limits_{x \to 0} {{{e^x} – 1} \over {\sqrt {x + 1} – 1}}\)
\(=\mathop {\lim }\limits_{x \to 0} {{{e^x} – 1} \over {\sqrt {x + 1} – 1}} = \mathop {\lim }\limits_{x \to 0} {{({e^x} – 1)(\sqrt {x + 1} + 1)} \over x}\)
Advertisements (Quảng cáo)
\( = \mathop {\lim }\limits_{x \to 0} {{{e^x} – 1} \over x}.(\sqrt {x + 1} + 1) = 2\)
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) \(\mathop {\lim }\limits_{x \to 0} {{\ln \left( {1 + 2x} \right)} \over {\tan x}} = \mathop {\lim }\limits_{x \to 0} {{{{\ln \left( {1 + 2x} \right)} \over {2x}}} \over {{{\tan x} \over x}}}.2 = {1 \over 1}.2 = 2\)