(nsin(1/n))^n^2在n趋近于无穷大时的极限
3个回答

n→∞lim(nsin1/n)^n²

=n→∞lim[(sin1/n)/(1/n)]^n²

=x→0lim[(sinx)/x)]^(1/x)²

=x→0lime^ln[(sinx)/x)]^(1/x)²

=x→0lime^[(1/x)²]ln[(sinx)/x)]^(1/x)²

=x→0lime^{[(1/x)²]*ln[(sinx)/x)]}

=x→0lime^{ln[(sinx)/x)]/x²}

=x→0lime^{[(x/sinx)*(xcosx-sinx)/x²]/2x}(罗比塔法则)

=x→0lime^{[(x/x)*(xcosx-x)/x²]/2x}(等量替换)

=x→0lime^{[(cosx-1)/2x²]}

=x→0lime^{[(-sinx)/4x]}(罗比塔法则)

=x→0lime^{[(-x)/4x]}

=x→0lime^{[-1/4]}(等量替换)

=e^(-1/4)