Limit I don't know how to start solving. This is the limit:
$$\lim_{n\to\infty}\frac{2^{n-1}-4^n}{3^n\sin n + 4^{n+2}}$$
I have a solution and the steps and I still haven't understood how it's done, here's the proposed solution:
$$\lim_{n\to\infty}\frac{2^{n-1}-4^n}{3^n\sin n + 4^{n+2}}=\lim_{n\to\infty}\frac{\frac12(\frac24)^n-1}{(\frac34)^n\sin n+16}=-\frac1{16}$$
WolframAlpha says this is correct but I haven't understood from where did all the fractions come from...

What you have to do is divide both the numerator and the denominator by $4^n$:
$$\frac{2^{n-1}-4^n}{4^n} = \frac{0.5\cdot2^n - 4^n}{4^n} = 0.5\left(\frac12\right)^n - 1$$
$$\frac{3^n\sin n + 4^{n+2}}{4^n} = \left(\frac34\right)^n + 4^2$$
$\left(\frac12\right)^n$ goes to $0$ and so does $\left(\frac34\right)^n$ when $n \to \infty$. Therefore the only terms that do not go to $0$ are the $1$ and the $16$.
