Question

limx-- 0 xtan2x - 2x tanx /((1- cos2x)2

Answer 1

Answer:

1/2

Step-by-step explanation:

limx-> 0 (xtan2x - 2x tanx)/((1- cos2x)²

Using Mclaurin's expansion for tan(x) = x + x³/3 +2/15x⁵+..., we get

xtan2x - 2x tanx = x[(2x) + (2x)³/3 + 2/15(2x)⁵+...]

-2x[x + x³/3 +2/15x⁵+....]

= 2x⁴ + 4x⁶ +....---(1)

Also, 1- cos2x = 2sin²x

limx->0 sinx/x =1,----(2),

limx-> 0 (xtan2x - 2x tanx)/((1- cos2x)²

Using (1) and (2),

=limx-> 0 (2x⁴ + 4x⁶ +...)/(2sin²x)²

Dividing Numerator and Denominator by x⁴ we get

=limx-> 0 (2 + 4x²+..)/4(sin⁴x/x⁴)

=limx-> 0 (2 + 4x²+..)/limx-> 04(sin⁴x/x⁴)

=1/2 .