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I need to find the following limit : $$\lim_{x\to \infty}\left(2^x\cdot \sin\left(\frac{b}{2^x}\right)\right)$$

I have tried it but I keep getting stuck, so any help would be helpful! Thank you!

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    $\begingroup$ Try writing it as $\lim_{x\to\infty}\frac{\sin(b\cdot 2^{-x})}{2^{-x}}$ and applying L'Hospital's rule. You may also find a change of variables useful. $\endgroup$ – DMcMor Mar 20 '17 at 20:05
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By substituting $u = 2^x$, this is $$\lim_{u \to \infty} u \sin \left(\frac{b}{u} \right)$$

You can do this by L'Hôpital: $$\lim_{u \to \infty} \frac{\sin \left(\frac{b}{u}\right)}{1/u}$$ which takes us to $$\lim_{u \to \infty} b \cos \left( \frac{b}{u} \right)$$ which I'm sure you can finish.

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  • $\begingroup$ Thank you so much I got it now!! $\endgroup$ – Zeemz97 Mar 20 '17 at 21:26
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HINT: $$\lim_{x\to\infty}2^x\sin\left(\frac{b}{2^x}\right)=\lim_{x\to\infty}b\frac{\sin(\frac{b}{2^x})}{\frac{b}{2^x}}$$

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