Tìm các giới hạn sau:
a) \(\mathop {\lim }\limits_{_{x \to 0}} \frac{{\ln (1 + 3x)}}{x}\)
b) \(\mathop {\lim }\limits_{_{x \to 0}} \frac{{\ln (1 + {x^2})}}{x}\)
a) \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + 3x} \right)}}{x} = 3.\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + 3x} \right)}}{{3x}} = 3\)
b) Vì \(\mathop {\lim }\limits_{_{x \to 0}} \frac{{\ln (1 + {x^2})}}{{{x^2}}} = 1\) nên
\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + {x^2}} \right)}}{x}\\
= \mathop {\lim }\limits_{x \to 0} x\frac{{\ln \left( {1 + {x^2}} \right)}}{{{x^2}}} = 0.1 = 0
\end{array}\)
-- Mod Toán 12
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