Question Number 180359 by mathlove last updated on 11/Nov/22
Answered by Frix last updated on 11/Nov/22
![certainly you do not mean x! which is defined for x∈N ⇒ no limit exists but you mean x!=Γ(x+1) we know Γ(1)=1 let Γ(x+1)=f(x) we also know ((√y))^(√y) =y^((√y)/2) lim_(x→0) ((f(x)−(√(f(x))))/(f(x)^(f(x)) −f(x)^((√(f(x)))/2) )) = =lim_(x→0) (((d[f(x)−(√(f(x)))])/dx)/((d[f(x)^(f(x)) −f(x)^((√(f(x)))/2) ])/dx)) = =lim_(x→0) ((((d[f(x)])/dx)(1−(1/(2(√(f(x)))))))/(((d[f(x))])/dx)(f(x)^(f(x)) (1+ln f(x))−((f(x)^(((√(f(x)))−1)/2) (2+ln f(x)))/4)))) = =lim_(x→0) ((2(2(√(f(x)))−1))/(4f(x)^(f(x)+(1/2)) (1+ln f(x))−f(x)^((√(f(x)))/2) (2+ln f(x)))) = =1](https://www.tinkutara.com/question/Q180360.png)
$$\mathrm{certainly}\:\mathrm{you}\:\mathrm{do}\:\mathrm{not}\:\mathrm{mean}\:{x}!\:\mathrm{which}\:\mathrm{is} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{N}\:\Rightarrow\:\mathrm{no}\:\mathrm{limit}\:\mathrm{exists} \\ $$$$\mathrm{but}\:\mathrm{you}\:\mathrm{mean}\:{x}!=\Gamma\left({x}+\mathrm{1}\right) \\ $$$$\mathrm{we}\:\mathrm{know}\:\Gamma\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{let}\:\Gamma\left({x}+\mathrm{1}\right)={f}\left({x}\right) \\ $$$$\mathrm{we}\:\mathrm{also}\:\mathrm{know}\:\left(\sqrt{{y}}\right)^{\sqrt{{y}}} ={y}^{\frac{\sqrt{{y}}}{\mathrm{2}}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left({x}\right)−\sqrt{{f}\left({x}\right)}}{{f}\left({x}\right)^{{f}\left({x}\right)} −{f}\left({x}\right)^{\frac{\sqrt{{f}\left({x}\right)}}{\mathrm{2}}} }\:= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}\left[{f}\left({x}\right)−\sqrt{{f}\left({x}\right)}\right]}{{dx}}}{\frac{{d}\left[{f}\left({x}\right)^{{f}\left({x}\right)} −{f}\left({x}\right)^{\frac{\sqrt{{f}\left({x}\right)}}{\mathrm{2}}} \right]}{{dx}}}\:= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}\left[{f}\left({x}\right)\right]}{{dx}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{f}\left({x}\right)}}\right)}{\frac{\left.{d}\left[{f}\left({x}\right)\right)\right]}{{dx}}\left({f}\left({x}\right)^{{f}\left({x}\right)} \left(\mathrm{1}+\mathrm{ln}\:{f}\left({x}\right)\right)−\frac{{f}\left({x}\right)^{\frac{\sqrt{{f}\left({x}\right)}−\mathrm{1}}{\mathrm{2}}} \left(\mathrm{2}+\mathrm{ln}\:{f}\left({x}\right)\right)}{\mathrm{4}}\right)}\:= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\mathrm{2}\sqrt{{f}\left({x}\right)}−\mathrm{1}\right)}{\mathrm{4}{f}\left({x}\right)^{{f}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}+\mathrm{ln}\:{f}\left({x}\right)\right)−{f}\left({x}\right)^{\frac{\sqrt{{f}\left({x}\right)}}{\mathrm{2}}} \left(\mathrm{2}+\mathrm{ln}\:{f}\left({x}\right)\right)}\:= \\ $$$$=\mathrm{1} \\ $$