Prove that limits can be used for asymptotic analysis True or false:
If f(n)=$\Theta$(g(n)), then $$\lim_{n\rightarrow \infty}\frac{f(n)}{g(n)}$$ 
exists and is equal to some real number.
I'm not sure what needs to be done to demonstrate this. I do understand the laws of limits and properties of limits. 
 A: As an expansion on the perfect comments of Daniel and Crostul. Consider the function:
$$\begin{cases}f:\Bbb N\to\Bbb N \\
2k\mapsto 3k \\ 2k+1\mapsto 2k\end{cases}$$
$$\begin{cases} g: \Bbb N\to\Bbb N \\ n\mapsto n\end{cases}$$
This sequence has $f=\Theta(g)$ by definiting, because
$$|f(n)|\le 2|g(n)|$$ for every $n$, and similarly for every $n$.
$$|g(n)|\le 2|f(n)|$$
Now, for a limit to exist, we need that the value of ${f(n)\over g(n)}$ approaches some number as $n\to\infty$.
However, if $n$ is even, the ratio
$${f(2k)\over g(2k)}={3k\over 2k}={3\over 2}$$
and if $n$ is odd, the ratio is
$${f(2k+1)\over g(2k+1)}={2k\over 2k+1}\stackrel{k\to\infty}{\longrightarrow} 1$$
Formally this is written
$$\limsup_{n\to\infty} {f\over g} = {3\over 2} \\ \liminf_{n\to\infty} {f\over g} = 1$$
because the "biggest" limit you could get is ${3\over 2}$ and the "smallest" is $1$, but you don't need to know the terminology for that.
In either case, the limit does not exist for the same reason
$$\lim_{x\to\infty}\sin x$$
does not:  it never approaches anything, it just bounces back and forth between two values.

Addendum: I chose this function because it actually has non-trivial computations for limits. If you want the "pocket" answer, it's even easier:
Let $f(n) = 1$ no matter what $n$ is and let $g(n) = (-1)^n$ so that
$$|f(n)|\le |g(n)|\le |f(n)|$$
immediately verifying $f(n)=\Theta(g(n))$, but the limit of the ratio bounces back and forth between $\pm 1$ forever.
