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Question Number 184159 by Davidtim last updated on 03/Jan/23
prove that 0^0 =1
$${prove}\:{that}\:\mathrm{0}^{\mathrm{0}} =\mathrm{1} \\ $$
Commented by Frix last updated on 03/Jan/23
Prove that the moon is a star.
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{moon}\:\mathrm{is}\:\mathrm{a}\:\mathrm{star}. \\ $$
Answered by cortano2 last updated on 03/Jan/23
let 0^n =x  0^0 =0^(n−n) =(0^n /0^n )=(x/x)=1
$${let}\:\mathrm{0}^{{n}} ={x} \\ $$$$\mathrm{0}^{\mathrm{0}} =\mathrm{0}^{{n}−{n}} =\frac{\mathrm{0}^{{n}} }{\mathrm{0}^{{n}} }=\frac{{x}}{{x}}=\mathrm{1} \\ $$
Commented by Frix last updated on 03/Jan/23
(0/0) is not defined.  Let f(0)=g(0)=0  lim_(x→0)  ((f(x))/(g(x))) =lim_(x→0)  ((f′(x))/(g′(x))) which can be any real number  i.e.  lim_(x→0)  (x^2 /(sin x)) =lim_(x→0)  ((2x)/(cos x)) =0  lim_(x→0)  (x/x^2 ) =lim_(x→0)  (1/(2x)) not defined  lim_(x→0)  ((xcos x)/(x+sin x)) =lim_(x→0)  ((cos x −xsin x)/(1+cos x)) =(1/2)
$$\frac{\mathrm{0}}{\mathrm{0}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}. \\ $$$$\mathrm{Let}\:{f}\left(\mathrm{0}\right)={g}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left({x}\right)}{{g}\left({x}\right)}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}'\left({x}\right)}{{g}'\left({x}\right)}\:\mathrm{which}\:\mathrm{can}\:\mathrm{be}\:\mathrm{any}\:\mathrm{real}\:\mathrm{number} \\ $$$$\mathrm{i}.\mathrm{e}. \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} }{\mathrm{sin}\:{x}}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}}{\mathrm{cos}\:{x}}\:=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}}{{x}^{\mathrm{2}} }\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{2}{x}}\:\mathrm{not}\:\mathrm{defined} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\mathrm{cos}\:{x}}{{x}+\mathrm{sin}\:{x}}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:{x}\:−{x}\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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