Question

lim xtan4x/1-cos4x x-0

Answer 1

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
$\lim_{x \to \zero} (4xsec^24x +tan4x)/4sin4x$
=0 
as tan0° =0 and sin0° =0