Bài 53. Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{x \to 0} {{\ln \left( {1 + 3x} \right)} \over x}\)
b) \(\mathop {\lim }\limits_{x \to 0} {{\ln \left( {1 + {x^2}} \right)} \over x}\)
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a) \(\mathop {\lim }\limits_{x \to 0} {{\ln \left( {1 + 3x} \right)} \over x} = 3.\mathop {\lim }\limits_{x \to 0} {{\ln \left( {1 + 3x} \right)} \over {3x}} = 3\).
b) Vì \(\mathop {\lim }\limits_{x \to 0} {{\ln \left( {1 + {x^2}} \right)} \over {{x^2}}} = 1\) nên \(\mathop {\lim }\limits_{x \to 0} {{\ln \left( {1 + {x^2}} \right)} \over x} = \mathop {\lim }\limits_{x \to 0} x{{\ln \left( {1 + {x^2}} \right)} \over {{x^2}}} = 0.1 = 0\).