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Question Number 120131 by huotpat last updated on 29/Oct/20

Commented by Dwaipayan Shikari last updated on 29/Oct/20

lim_(x→3) ((2^x −2^3 )/(sin(πx)))=2^3 ((2^(x−3) −1)/(sin(3π−πx))) =2^3 .((2^(x−3) −1)/(π(3−x)))=−(2^3 /π).((2^(x−3) −1)/(x−3))=−(8/π).log(2) sin(3π−θ)=sinθ sin(3π−3x)→3π−3x

$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\mathrm{2}^{{x}} −\mathrm{2}^{\mathrm{3}} }{{sin}\left(\pi{x}\right)}=\mathrm{2}^{\mathrm{3}} \frac{\mathrm{2}^{{x}−\mathrm{3}} −\mathrm{1}}{{sin}\left(\mathrm{3}\pi−\pi{x}\right)} \\ $$$$=\mathrm{2}^{\mathrm{3}} .\frac{\mathrm{2}^{{x}−\mathrm{3}} −\mathrm{1}}{\pi\left(\mathrm{3}−{x}\right)}=−\frac{\mathrm{2}^{\mathrm{3}} }{\pi}.\frac{\mathrm{2}^{{x}−\mathrm{3}} −\mathrm{1}}{{x}−\mathrm{3}}=−\frac{\mathrm{8}}{\pi}.{log}\left(\mathrm{2}\right) \\ $$$${sin}\left(\mathrm{3}\pi−\theta\right)={sin}\theta \\ $$$${sin}\left(\mathrm{3}\pi−\mathrm{3}{x}\right)\rightarrow\mathrm{3}\pi−\mathrm{3}{x} \\ $$

Answered by Bird last updated on 29/Oct/20

f(x)=((2^x −8)/(sin(πx))) ⇒ f(x)=_(x−3=t) ((2^(t+3) −8)/(sin(π(t+3)))) =((8(2^t −1))/(sin(πt+π))) =−8((2^t −1)/(sin(πt))) 2^t −1 =e^(tln2) −1 ∼1+tln2−1=tln2 ⇒ f(3+t)∼−8.((tln2)/(sin(πt))) =−((8ln(2))/π)×((πt)/(sin(πt))) →−((8ln(2))/π) ⇒lim_(t→o) f(3+t) =−((8ln2)/π) =lim_(x→3) f(x)

$${f}\left({x}\right)=\frac{\mathrm{2}^{{x}} −\mathrm{8}}{{sin}\left(\pi{x}\right)}\:\Rightarrow \\ $$$${f}\left({x}\right)=_{{x}−\mathrm{3}={t}} \:\:\:\frac{\mathrm{2}^{{t}+\mathrm{3}} −\mathrm{8}}{{sin}\left(\pi\left({t}+\mathrm{3}\right)\right)} \\ $$$$=\frac{\mathrm{8}\left(\mathrm{2}^{{t}} −\mathrm{1}\right)}{{sin}\left(\pi{t}+\pi\right)}\:=−\mathrm{8}\frac{\mathrm{2}^{{t}} −\mathrm{1}}{{sin}\left(\pi{t}\right)} \\ $$$$\mathrm{2}^{{t}} −\mathrm{1}\:={e}^{{tln}\mathrm{2}} −\mathrm{1}\:\sim\mathrm{1}+{tln}\mathrm{2}−\mathrm{1}={tln}\mathrm{2}\:\Rightarrow \\ $$$${f}\left(\mathrm{3}+{t}\right)\sim−\mathrm{8}.\frac{{tln}\mathrm{2}}{{sin}\left(\pi{t}\right)} \\ $$$$=−\frac{\mathrm{8}{ln}\left(\mathrm{2}\right)}{\pi}×\frac{\pi{t}}{{sin}\left(\pi{t}\right)}\:\rightarrow−\frac{\mathrm{8}{ln}\left(\mathrm{2}\right)}{\pi} \\ $$$$\Rightarrow{lim}_{{t}\rightarrow{o}} {f}\left(\mathrm{3}+{t}\right)\:=−\frac{\mathrm{8}{ln}\mathrm{2}}{\pi} \\ $$$$={lim}_{{x}\rightarrow\mathrm{3}} {f}\left({x}\right) \\ $$

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