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For n belong to N fn(x)=tan x/2 (1+sec x)(1+sec 2x)(1+sec4x).........(1+sec 2^n x) Then lim fn(x)/2x is equal to

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Solution:  We have , nin N

                 f_n(x)=	an fracx2(1+sec x)(1+sec 2x)(1+sec 4x)........(1+sec 2^xx)

      \ \ Rightarrow f_n(x)=	an fracx2(frac1+cos xcos x)(1+sec 2x)(1+sec 4x)........(1+sec 2^nx)\ \ =	an x (frac1+cos 2xcos 2x)(1+sec 4x).........(1+sec 2^nx)\ \ =	an 2x (1+sec 2^2x)........(1+sec 2^nx)\ \ =	an 2^n-1x(1+sec 2^nx)=	an 2^nx  

Now ,       lim_x	o 0fracf_n(x)2x=lim _x	o 0frac	an 2^nx2^nxcdot 2^n-1=2^n-1

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Deependra Verma

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