题目
1.当x→∞时,sin^2(1)/(x)与(1)/(x^k)为等价无穷小,则k=
1.当x→∞时,$\sin^{2}\frac{1}{x}$与$\frac{1}{x^{k}}$为等价无穷小,则k=
题目解答
答案
当 $x \to \infty$ 时,$\frac{1}{x} \to 0$,根据等价无穷小性质 $\sin u \sim u$(当 $u \to 0$),有:
$\sin^2 \frac{1}{x} \sim \left(\frac{1}{x}\right)^2 = \frac{1}{x^2}$
为使 $\sin^2 \frac{1}{x}$ 与 $\frac{1}{x^k}$ 等价,需满足:
$\lim_{x \to \infty} \frac{\sin^2 \frac{1}{x}}{\frac{1}{x^k}} = 1 \quad \Rightarrow \quad \lim_{x \to \infty} \frac{\frac{1}{x^2}}{\frac{1}{x^k}} = 1 \quad \Rightarrow \quad \lim_{x \to \infty} x^{k-2} = 1$
由极限性质,$k-2 = 0$,解得 $k = 2$。
答案: $k = 2$