Question Number 24755 by Anoop kumar last updated on 25/Nov/17
![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) .](Q24755.png)
$${Given} \\ $$$${f}\left({x}\right)\:=\underset{{x}=\mathrm{1}} {\overset{{n}} {\sum}}{tan}\left(\frac{{x}}{\mathrm{2}^{{r}} }\right).{sec}\left(\frac{{x}}{\mathrm{2}^{{r}−\mathrm{1}} }\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:{where}\:{r}\:{and}\:{n}\:\varepsilon{N} \\ $$$${g}\left({x}\right)\:=\underset{{n}\rightarrow\propto} {\mathrm{li}{m}}\:\:\frac{{ln}\left({f}\left({x}\right)+{tan}\frac{{x}}{\mathrm{2}^{{n}} }\right)\:−\left({f}\left({x}\right)+{tan}\frac{{x}}{\mathrm{2}^{{n}} }\right).\left[{sin}\left({tan}\frac{{x}}{\mathrm{2}}\right)\right.}{\mathrm{1}+\left({f}\left({x}\right)\:\:+\:\:{tan}\frac{{x}}{\mathrm{2}^{{n}} }\right)^{{n}} }\:\:=\:{k} \\ $$$${for}\:{x}\:=\frac{\pi}{\mathrm{4}}\:\:{and}\:{the}\:{domain}\:{of} \\ $$$${g}\left({x}\right)\:{is}\:\left(\mathrm{0}\:,\frac{\pi}{\mathrm{2}}\right) \\ $$$${where}\:\left[.\right]\:{denotes}\:{the}\:{g}.{i}.\mathrm{f} \\ $$$${Find}\:{the}\:{value}\:{of}\:{k},\:{if}\:{possible}\: \\ $$$${so}\:{that}\:{g}\left({x}\right)\:{is}\:{continuous}\:{at}\: \\ $$$${x}\:=\frac{\pi}{\mathrm{4}}\:. \\ $$$$ \\ $$$$ \\ $$