Evaluate `lim_(x to 0) (log_(tan^(2)x)(tan^(2)2x).`

Explanation

`L=underset(xto0)lim(log(tan^(2)2x))/(log(tan^(2)x))" "`(`(oo)/(oo)` form)
Using L'Hospital's rule, we have
`L=underset(xto0)lim(((1)/(tan^(2)2x)2tan2xsec^(2)2x)xx2)/((1)/(tan^(2)x)2tanx.sec^(2)x)`
`=underset(xto0)lim(2((1)/(sin2xcos2x)))/(((1)/(sinxcosx)))=underset(xto0)lim(((1)/(sin2xcos2x)))/(((1)/(sin2x)))`
`=underset(xto0)lim(1)/(cos^(2)x)=1`