Evaluating the limit: $\lim _{x\to \infty }\left(2^x\sin\left(\frac{b}{2^x}\right)\right)$ 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!

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.
