Question Number 4025 by 123456 last updated on 26/Dec/15
![f_(n+1) (x)=x^(f_n (x)) g_(n+1) (x)=[g_n (x)]^x h_(n+1) (x)=[h_n (x)]^(h_n (x)) f_0 (x)=g_0 (x)=h_0 (x)=x if f(x)=lim_(n→∞) f_n (x) g(x)=lim_(n→∞) g_n (x) h(x)=lim_(n→∞) h_n (x) does f(x)=g(x)=h(x)?](Q4025.png)
$${f}_{{n}+\mathrm{1}} \left({x}\right)={x}^{{f}_{{n}} \left({x}\right)} \\ $$$${g}_{{n}+\mathrm{1}} \left({x}\right)=\left[{g}_{{n}} \left({x}\right)\right]^{{x}} \\ $$$${h}_{{n}+\mathrm{1}} \left({x}\right)=\left[{h}_{{n}} \left({x}\right)\right]^{{h}_{{n}} \left({x}\right)} \\ $$$${f}_{\mathrm{0}} \left({x}\right)={g}_{\mathrm{0}} \left({x}\right)={h}_{\mathrm{0}} \left({x}\right)={x} \\ $$$$\mathrm{if} \\ $$$${f}\left({x}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}_{{n}} \left({x}\right) \\ $$$${g}\left({x}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{g}_{{n}} \left({x}\right) \\ $$$${h}\left({x}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{h}_{{n}} \left({x}\right) \\ $$$$\mathrm{does} \\ $$$${f}\left({x}\right)={g}\left({x}\right)={h}\left({x}\right)? \\ $$
Commented by prakash jain last updated on 27/Dec/15
![I think limits exists only for x∈(0,1]](Q4051.png)
$$\mathrm{I}\:\mathrm{think}\:\mathrm{limits}\:\mathrm{exists}\:\mathrm{only}\:\mathrm{for}\:{x}\in\left(\mathrm{0},\mathrm{1}\right] \\ $$
Answered by Yozzii last updated on 27/Dec/15
![For this attempt, x is assumed given. (1)f_(n+1) (x)=x^(f_n (x)) . f(x)=lim_(n→∞) f_n (x) [This indicates a limit exists.] ∴ f(x)=x^(f(x)) (unsure how to solve...) (2)g_(n+1) (x)=(g_n (x))^x . g(x)=lim_(n→∞) g_n (x) ⇒g(x)=(g(x))^x (g(x))^x −g(x)=0 g(x)((g(x))^(x−1) −1)=0 ⇒g(x)=0 ∨ (g(x))^(x−1) =1 If x≠1⇒ g(x)=1^(1/(x−1)) ⇒g(x)=1 g(x)=0 ∨ g(x)=1 (3)h(x)=lim_(n→∞) h_n (x) h_(n+1) (x)=[h_n (x)]^(h_n (x)) ∴ h(x)={h(x)}^(h(x)) {h(x)}^(h(x)) −h(x)=0 h(x)[{h(x)}^(h(x)−1) −1]=0 ⇒h(x)=0 ∨ {h(x)}^(h(x)−1) −1=0 {h(x)}^(h(x)−1) =1 (∗) (∗) is true if (i)h(x)−1=0⇒h(x)=1 (exponent) (ii)h(x)=1 (base) (iii)h(x)=−1⇒ (−1)^(−2) =(1/((−1)^2 ))=1 h(x)=0 ∨ h(x)=1 ∨ h(x)=−1 ∴ ∃g(x),h(x)⇒g(x)=h(x).](Q4030.png)
$${For}\:{this}\:{attempt},\:{x}\:{is}\:{assumed}\:{given}. \\ $$$$\left(\mathrm{1}\right){f}_{{n}+\mathrm{1}} \left({x}\right)={x}^{{f}_{{n}} \left({x}\right)} . \\ $$$${f}\left({x}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}_{{n}} \left({x}\right)\:\:\left[{This}\:{indicates}\:{a}\:{limit}\:{exists}.\right] \\ $$$$\therefore\:{f}\left({x}\right)={x}^{{f}\left({x}\right)} \:\left({unsure}\:{how}\:{to}\:{solve}...\right) \\ $$$$ \\ $$$$\left(\mathrm{2}\right){g}_{{n}+\mathrm{1}} \left({x}\right)=\left({g}_{{n}} \left({x}\right)\right)^{{x}} . \\ $$$${g}\left({x}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{g}_{{n}} \left({x}\right) \\ $$$$\Rightarrow{g}\left({x}\right)=\left({g}\left({x}\right)\right)^{{x}} \\ $$$$\left({g}\left({x}\right)\right)^{{x}} −{g}\left({x}\right)=\mathrm{0} \\ $$$${g}\left({x}\right)\left(\left({g}\left({x}\right)\right)^{{x}−\mathrm{1}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{g}\left({x}\right)=\mathrm{0}\:\vee\:\left({g}\left({x}\right)\right)^{{x}−\mathrm{1}} =\mathrm{1} \\ $$$${If}\:{x}\neq\mathrm{1}\Rightarrow\:{g}\left({x}\right)=\mathrm{1}^{\frac{\mathrm{1}}{{x}−\mathrm{1}}} \Rightarrow{g}\left({x}\right)=\mathrm{1} \\ $$$${g}\left({x}\right)=\mathrm{0}\:\vee\:{g}\left({x}\right)=\mathrm{1} \\ $$$$ \\ $$$$\left(\mathrm{3}\right){h}\left({x}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{h}_{{n}} \left({x}\right) \\ $$$${h}_{{n}+\mathrm{1}} \left({x}\right)=\left[{h}_{{n}} \left({x}\right)\right]^{{h}_{{n}} \left({x}\right)} \\ $$$$\therefore\:{h}\left({x}\right)=\left\{{h}\left({x}\right)\right\}^{{h}\left({x}\right)} \\ $$$$\left\{{h}\left({x}\right)\right\}^{{h}\left({x}\right)} −{h}\left({x}\right)=\mathrm{0} \\ $$$${h}\left({x}\right)\left[\left\{{h}\left({x}\right)\right\}^{{h}\left({x}\right)−\mathrm{1}} −\mathrm{1}\right]=\mathrm{0} \\ $$$$\Rightarrow{h}\left({x}\right)=\mathrm{0}\:\vee\:\left\{{h}\left({x}\right)\right\}^{{h}\left({x}\right)−\mathrm{1}} −\mathrm{1}=\mathrm{0} \\ $$$$\left\{{h}\left({x}\right)\right\}^{{h}\left({x}\right)−\mathrm{1}} =\mathrm{1}\:\:\left(\ast\right) \\ $$$$\left(\ast\right)\:{is}\:{true}\:{if}\:\left({i}\right){h}\left({x}\right)−\mathrm{1}=\mathrm{0}\Rightarrow{h}\left({x}\right)=\mathrm{1}\:\left({exponent}\right) \\ $$$$\left({ii}\right){h}\left({x}\right)=\mathrm{1}\:\left({base}\right) \\ $$$$\left({iii}\right){h}\left({x}\right)=−\mathrm{1}\Rightarrow\:\left(−\mathrm{1}\right)^{−\mathrm{2}} =\frac{\mathrm{1}}{\left(−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{1} \\ $$$${h}\left({x}\right)=\mathrm{0}\:\vee\:{h}\left({x}\right)=\mathrm{1}\:\vee\:{h}\left({x}\right)=−\mathrm{1} \\ $$$$\therefore\:\exists{g}\left({x}\right),{h}\left({x}\right)\Rightarrow{g}\left({x}\right)={h}\left({x}\right). \\ $$$$ \\ $$