f ( x ) = tan ( 2 x ) ⋅ sec ( x ) + tan ( 2 2 x ) ⋅ sec ( 2 x ) + … + tan ( 2 n x ) ⋅ sec ( 2 n − 1 x ) g ( x ) = f ( x ) + tan ( 2 n x )
where x ∈ ( − 2 π , 2 π ) and n ∈ N
Evaluate the limit : x → 0 lim ( x g ( x ) ) x 3 1
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G(x)=tan(x). For proof check the previous problem (bcs I hate latex,sorry)so apply the limit u will that limit as e^infinity so not possible