求极限lim(x→0){(2x-sin2x)/xsin^2x}
人气:199 ℃ 时间:2020-02-03 13:27:42
解答
lim(x→0){(2x-sin2x)/(x*sin^2x)}
= lim(x→0){(2x-sin2x)/(x*x^2*(sin^2x/x^2))}
= lim(x→0){(2x-sin2x)/(x*x^2) * lim(x→0){(sin^2x/x^2)}
= lim(x→0){(2-2cos2x)/(3x^2)
= lim(x→0){(4sin2x)/(6x)
= lim(x→0){(8cos2x)/(6)
= 4/3
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