Prove that $\frac{n^2}{2} - 3n = \Theta(n^2)$ The question is,

Prove that $\frac{n^2}{2} - 3n = \Theta(n^2)$.

I understand that to do this I must determine positive constants $c_1$, $c_2$, and $n_0$ such that
$$c_1n^2 \leq \frac{n^2}{2} - 3n  \leq c_2n^2$$
I simplified by dividing by $n^2$ which left
$$c_1 \leq \frac{1}{2} - \frac{3}{n} \leq c_2$$
The book I am following along with says "We can make the right-hand inequality hold for any value of $n \geq 1$ by choosing any constant $c_2 \geq \frac{1}{2}$. Likewise, we can make the left-hand inequality hold for any value of $n \geq 7$ by choosing any constant $c_1 \leq \frac{1}{14}$."
I understand that there are other choices for the constants but I'm not sure of a method for determining them. What general procedure should I use for determining these other than guess and check? 
 A: What you did so far was good. Then, to determine an exact choice for the constants $c_1$ and $c_2$, you should think carefully about the expression, in this case
$$
\frac{1}{2} - \frac{3}{n}.
$$
As $n$ gets larger, what does this sequence do? Well, it gets closer and closer to $\frac12$. And it only increases; it gets closer to $\frac12$ from below.
That means we can pick the upper bound $c_2 = \frac12$. And since it increases, we can pick the lower bound $c_1$ to be the $n$th term for some suitably large $n$. In the book, they picked $n = 7$ to get $\frac{1}{2} - \frac{3}{7} = \frac{1}{14}$ as the lower bound.
A: Your task doesn't really require you to specify the constants. One approach for the solution would be to state that the sequence
$$ \frac{1}{2} - \frac{3}{n} $$
converges to the constant $ \frac{1}{2} $. This means that for any $ \varepsilon > 0 $ there exists $ N(\varepsilon)$ such that $ \frac{1}{2} - \varepsilon < \frac{1}{2} - \frac{3}{n} < \frac{1}{2} + \varepsilon $ for $n > N(\varepsilon) $, which gives you the two constans for all sufficiently large $ n $ (i.e. larger than $ N(\varepsilon) $)
