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Tính các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to 9} {\log _3}x\)
b) \(\mathop {\lim }\limits_{x \to 0} {{\ln \left( {4x + 1} \right)} \over x}\)
c) \(\mathop {\lim }\limits_{x \to 0} {{\ln \left( {3x + 1} \right) – \ln \left( {2x + 1} \right)} \over x}\) d) \(\mathop {\lim }\limits_{x \to 0} {{\ln \left( {1 + 3x} \right)} \over {\sin 2x}}\)
Hướng dẫn: d) Vận dụng công thức \(\mathop {\lim }\limits_{x \to 0} {{\sin x} \over x} = 1\)
Giải
a) \(\mathop {\lim }\limits_{x \to 9} {\log _3}x={\log _3}9 = 2\)
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b) \(\mathop {\lim }\limits_{x \to 0} {{\ln \left( {4x + 1} \right)} \over x}\)
\(=\mathop {\lim }\limits_{x \to 0} 4.{{\ln \left( {4x + 1} \right)} \over {4x}}=4.1=4\)
c) \(\mathop {\lim }\limits_{x \to 0} {{\ln \left( {3x + 1} \right) – \ln \left( {2x + 1} \right)} \over x} \)
\(= \mathop {\lim }\limits_{x \to 0} {{\ln \left( {3x + 1} \right)} \over {3x}}.3 – \mathop {\lim }\limits_{x \to 0} {{\ln \left( {2x + 1} \right)} \over {2x}}.2 = 3 – 2 = 1\)
d) \(\mathop {\lim }\limits_{x \to 0} {{\ln \left( {1 + 3x} \right)} \over {\sin 2x}} = \mathop {\lim }\limits_{x \to 0} {{{{\ln \left( {1 + 3x} \right)} \over {3x}}} \over {{{\sin 2x} \over {2x}}}}.{3 \over 2} = {{\mathop {\lim }\limits_{x \to 0} {{\ln \left( {1 + 3x} \right)} \over {3x}}} \over {\mathop {\lim }\limits_{x \to 0} {{\sin 2x} \over {2x}}}}.{3 \over 2} = {3 \over 2}\)