Câu 2.74 trang 82 sách bài tập Giải tích 12 Nâng cao

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) \(\mathop {\lim }\limits_{x \to 9} {\log _3}x={\log _3}9 = 2\)                                   

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

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