Question

limx→0sin2x/tankx=1,k=多少？

Answer 1

要求limx→0sin2x/tan(kx)=1，需要找到一个常数k使得limx→0(sin2x)/(tan(kx))=1。根据三角恒等式，tan(2x)=2tan(x)×(1-tan^2(x))，可以得到：sin(2x) = 2sin(x)cos(x)tan(2x) = 2tan(x)/(1-tan^2(x))因此，limx→0(sin2x)/(tan(kx))= limx→0(2sin(x)cos(x))/(2tan(x)/(1-tan^2(x)))= limx→0(cos(x))/(tan(x)/(1-tan^2(x)))= limx→0(cos(x))/(sin(x)/cos(x))/(1-(sin(x)/cos(x))^2)= limx→0(cos^2(x))/(sin(x))(1+(sin^2(x)/cos^2(x)))= limx→0(cos^2(x))/(sin(x))(cos^2(x)+sin^2(x))= 1/k因此，要使limx→0(sin2x)/(tan(kx))=1，k应该等于1。