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As  \lim_{x \to \0} (xtan4x)/(1 -cos4x) = 0/0 form ( x tends to 0)
so applying L' Hospital rule,i.e. differentiate the numerator and denominator separately and then take the limit.
so, (xtan4x)/(1 -cos4x) = (4xsec²4x +tan4x)/4sin4x
now taking limit
 \lim_{x \to \zero}  (4xsec^24x +tan4x)/4sin4x
as tan0° =0 and sin0° =0