The function f(x) =(sin2x)tan22x  is not defined at x=π/4 , then the value of f(π/4)  so that f  is continuous at x=π/4  is

f  is continuous at x=π/4  if limx→π/4 f(x)=f(π/4) Now L=limx→π/4(sin2x)tan22x⇒ logL=limx→π/4tan22xlogsin2x                =limx→π/4logsin2xcot22x(00 form using L'Hospital rule)                =limx→π/42cot2x−2cot2xcosec22x.2                =–12∴    L=e-1 2  ⇒  f(π4)=e−1/2=1e .