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Question Number 46673 by Tawa1 last updated on 30/Oct/18

Commented by Smail last updated on 30/Oct/18

$$\alpha isaconstantyoushouldnotreplaceitwitht{2}^x.$$

Commented by Smail last updated on 30/Oct/18

$$Therightansweris$$$$\underset{t\to 0}{lim}\frac{sin\left(\alpha \right)}{\frac{\alpha }{t}sin\left(t\right)}=\frac{sin\left(\alpha \right)}{\alpha }$$

Commented by maxmathsup by imad last updated on 30/Oct/18

$$yesyouarerightidontcaretothispointwehave$$$$A\left(x\right)=\frac{sin\left(\alpha \right)}{2^{x}sin\left(\frac{\alpha }{2^x}\right)}=\frac{sin\left(\alpha \right)}{\alpha \frac{sin\left(\frac{\alpha }{2^x}\right)}{\frac{\alpha }{2^x}}}\to \frac{sin\left(\alpha \right)}{\alpha }(x\to +\mathrm{\infty })herewesuppose\alpha \ne k\pi$$$$if\alpha =k\pi A\left(x\right)=0\Rightarrow {lim}_{x\to +\mathrm{\infty }}A\left(x\right)=0$$$$$$

Commented by Tawa1 last updated on 30/Oct/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by maxmathsup by imad last updated on 30/Oct/18

$$youarewelcomesir.$$

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