evaluate without using series

lim     x tan 2x - 2x tan x

x -0    (1 - cos 2x )

Dear

Lt x_→o {xtan2x -2xtanx}/{1-cos2x}

= Lt x_→o {xtan2x -2xtanx}/{2sin²x}

=Lt x_→o {(tan2x -2tanx)/2x}/{(sinx)/x}²

=Lt x_→o {(tan2x -2tanx)/2x}           Lt x_→o sinx /x =1

= Lt x→o tan2x /2x    -2tanx/2x 

= 1 -1  =0

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