lim x→0 (ln(x^2+1))/xsin2

lim x→0 (ln(x^2+1))/xsin2x 的极限

1个回答

lim x→0 (ln(x^2+1))/xsin2x=lim(ln(1+x^2)/2x^2)/(sin2x/2x)
ln(1+x)~x
ln(1+x^2)~x^2
lim(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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