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Question-46673




Question Number 46673 by Tawa1 last updated on 30/Oct/18
Commented by Smail last updated on 30/Oct/18
α is a constant you should not replace it with t2^x .
$$\alpha\:{is}\:{a}\:{constant}\:{you}\:{should}\:{not}\:{replace}\:{it}\:{with}\:{t}\mathrm{2}^{{x}} .\: \\ $$
Commented by Smail last updated on 30/Oct/18
The right answer is   lim_(t→0) ((sin(α))/((α/t)sin(t)))=((sin(α))/α)
$${The}\:{right}\:{answer}\:{is}\: \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}\left(\alpha\right)}{\frac{\alpha}{{t}}{sin}\left({t}\right)}=\frac{{sin}\left(\alpha\right)}{\alpha} \\ $$
Commented by maxmathsup by imad last updated on 30/Oct/18
yes you are right i don t care to this point  w e have  A(x)=((sin(α))/(2^x  sin((α/2^x )))) = ((sin(α))/(α ((sin((α/2^(x ) )))/(α/2^x )))) →((sin(α))/α)  (x→+∞) here we suppose α≠kπ  if α=kπ   A(x) =0 ⇒lim_(x→+∞) A(x) =0
$${yes}\:{you}\:{are}\:{right}\:{i}\:{don}\:{t}\:{care}\:{to}\:{this}\:{point}\:\:{w}\:{e}\:{have} \\ $$$${A}\left({x}\right)=\frac{{sin}\left(\alpha\right)}{\mathrm{2}^{{x}} \:{sin}\left(\frac{\alpha}{\mathrm{2}^{{x}} }\right)}\:=\:\frac{{sin}\left(\alpha\right)}{\alpha\:\frac{{sin}\left(\frac{\alpha}{\mathrm{2}^{{x}\:} }\right)}{\frac{\alpha}{\mathrm{2}^{{x}} }}}\:\rightarrow\frac{{sin}\left(\alpha\right)}{\alpha}\:\:\left({x}\rightarrow+\infty\right)\:{here}\:{we}\:{suppose}\:\alpha\neq{k}\pi \\ $$$${if}\:\alpha={k}\pi\:\:\:{A}\left({x}\right)\:=\mathrm{0}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {A}\left({x}\right)\:=\mathrm{0} \\ $$$$ \\ $$
Commented by Tawa1 last updated on 30/Oct/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by maxmathsup by imad last updated on 30/Oct/18
you are welcome sir.
$${you}\:{are}\:{welcome}\:{sir}. \\ $$

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