Given f(x) =Σ_(x=1) ^n tan((x/2^r )).sec((x/2^(r−1) )) where r and n εN g(x) =lim_(n→∝) ((ln(f(x)+tan(x/2^n )) −(f(x)+tan(x/2^n )).[sin(tan(x/2)))/(1+(f(x) + tan(x/2^n ))^n )) = k for x =(π/4) and the domain of g(x) is (0 ,(π/2)) where [.] denotes the g.i.f Find the value of k, if possible so that g(x) is continuous at x =(π/4) .


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Question Number 24755 by Anoop kumar last updated on 25/Nov/17

$G i v e n$ $f (x) = \underset{x = 1}{\sum^{n}} t a n (\frac{x}{2^{r}}) . s e c (\frac{x}{2^{r - 1}})$ $w h e r e r a n d n ε N$ $g (x) = \underset{n \to\propto}{li m} \frac{l n (f (x) + t a n \frac{x}{2^{n}}) - (f (x) + t a n \frac{x}{2^{n}}) . [s i n (t a n \frac{x}{2})}{1 + {(f (x) + t a n \frac{x}{2^{n}})}^{n}} = k$ $f o r x = \frac{π}{4} a n d t h e d o m a i n o f$ $g (x) i s (0, \frac{π}{2})$ $w h e r e [.] d e n o t e s t h e g . i . f$ $F i n d t h e v a l u e o f k, i f p o s s i b l e$ $s o t h a t g (x) i s c o n t i n u o u s a t$ $x = \frac{π}{4} .$
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