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...
 A: Some intuition for how to come up with this solution:
As $n$ gets very large, $4^n$ is much much bigger than $2^n$. A common trick to "prove" this, or tease this fact out, is to divide the numerator and denominator by the highest order term. This is legal since it amounts to multiplying by 1, and $4^n$ is never 0 for any $n$. 
Doing this makes the result easier to see, as in your example. Once you divide by $4^n$ you end up with
$$
\frac{(\text{a bunch of stuff that goes to 0})-1}{(\text{a bunch of stuff that goes to 0})+4^{2}}\rightarrow-1/16
$$
as $n\rightarrow \infty$.
A: 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 $16$.
A: As often, the simplest is to use equivalents:
$2^{n-1}=_\infty o(4^n)$, hence $2^{n-1}- 4^n\sim_\infty- 4^n $. Similarly $\;3^n\sin n+4^{n+2}\sim_\infty4^{n+2}$, whence
$$\frac{2^{n-1}- 4^n}{3^n\sin n+4^{n+2}}\sim_\infty\frac{- 4^n}{4^{n+2}}=-\frac 1{4^2}.$$
