1/(sinx)^2 - 1/x^2在趋于0时为1/3,我用罗必塔法则算出怎么是0啊?

题目

1/(sinx)^2 - 1/x^2在趋于0时为1/3,我用罗必塔法则算出怎么是0啊?

答案

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以下是用洛必达的做法:
lim{x → 0} 1/(sin²(x))-1/x²
= lim{x → 0} (x²-sin²(x))/(x²sin²(x))
= (lim{x → 0} (x²-sin²(x))/x⁴)/(lim{x → 0} sin(x)/x)²
= lim{x → 0} (x²-sin²(x))/x⁴ (重要极限lim{x → 0} sin(x)/x = 1)
= lim{x → 0} (2x-2sin(x)cos(x))/(4x³) (0/0型,洛必达)
= lim{x → 0} (2-2cos²(x)+2sin²(x))/(12x²) (0/0型,洛必达)
= lim{x → 0} sin²(x)/(3x²)
= (lim{x → 0} sin(x)/x)²/3
= 1/3 (重要极限lim{x → 0} sin(x)/x = 1).
用Taylor展开也可以:
∵cos(t) = 1-t²/2+t⁴/24+o(t⁴),
∴sin²(x) = (1-cos(2x))/2 = (2x)²/4-(2x)⁴/48+o(x⁴) = x²-x⁴/3+o(x⁴),
∴x²-sin²(x) = x⁴/3+o(x⁴),即(x²-sin²(x))/x⁴ = 1/3+o(1) → 1/3.
再由上面已证lim{x → 0} 1/(sin²(x))-1/x² = lim{x → 0} (x²-sin²(x))/x⁴ = 1/3.