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