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lim x→0 (ln(x^2+1))/xsin2x 的极限
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人气:181 ℃ 时间:2020-01-28 22:40:42
解答
lim x→0 (ln(x^2+1))/xsin2x=lim(ln(1+x^2)/2x^2)/(sin2x/2x)ln(1+x)~xln(1+x^2)~x^2lim(ln(1+x^2)/2x^2)/(sin2x/2x)=(1/2)/1=1/2.即lim x→0 (ln(x^2+1))/xsin2x=1/2.
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