题目
11.已知lim_(x to 0) (ln(1+frac(f(x))/(x)))(2^x-1)=3,则lim_(x to 0) (f(x))/(sqrt(1+x^2))-1=_.
11.已知$\lim_{x \to 0} \frac{\ln(1+\frac{f(x)}{x})}{2^{x}-1}=3$,则$\lim_{x \to 0} \frac{f(x)}{\sqrt{1+x^{2}}-1}=\_.$
题目解答
答案
由题意,已知 $\lim_{x \to 0} \frac{\ln(1 + \frac{f(x)}{x})}{2^x - 1} = 3$。当 $x \to 0$ 时,利用等价无穷小替换:
$\ln(1 + \frac{f(x)}{x}) \sim \frac{f(x)}{x}, \quad 2^x - 1 \sim x \ln 2$
代入得:
$\lim_{x \to 0} \frac{\frac{f(x)}{x}}{x \ln 2} = 3 \implies \lim_{x \to 0} \frac{f(x)}{x^2} = 3 \ln 2$
即 $f(x) \sim 3 \ln 2 \cdot x^2$。
对于目标极限:
$\lim_{x \to 0} \frac{f(x)}{\sqrt{1 + x^2} - 1} \sim \lim_{x \to 0} \frac{3 \ln 2 \cdot x^2}{\frac{x^2}{2}} = 6 \ln 2$
(其中 $\sqrt{1 + x^2} - 1 \sim \frac{x^2}{2}$)。
答案: $\boxed{6 \ln 2}$