<s>[INST] Trying to find $c_{1}$ , $c_{2}$ for Big theta notation. Need to find $c_{1},$ $c_{2}$ and $n_{0}$.
\begin{equation}
c_{1}n^3 \leq \frac{n^3}{100} - 100n^2 - 100n + 3\leq c_{2}n^3
\end{equation}
\begin{equation}
c_{1} \leq \frac{1}{100} - \frac{100}{n} - \frac{100}{n^2} + \frac{3}{n^3} \leq c_{2}
\end{equation}
\begin{equation}
\frac{1}{100} - \frac{100}{n} - \frac{100}{n^2} + \frac{3}{n^3} \leq c_{2}
\end{equation}
This is what I have done so far, but I am currently stuck do I need to chose c2 which is positive and greater than $\frac{100}{n} + \frac{3}{n^3}$? And for c1 , I need a c1, which is >0. How do I find these constants? 
 [/INST] The first step is to find a suitable $n_0.$
You know that you need 
$$
0 < c_1 \leq \frac{1}{100} - \frac{100}{n} - \frac{100}{n^2} + \frac{3}{n^3} \leq c_2.
$$
Ignore the $c_1$ and $c_2$ for a moment, and you may see that you need
$$
0 < \frac{1}{100} - \frac{100}{n} - \frac{100}{n^2} + \frac{3}{n^3}.
$$
This must be true for every $n > n_0.$ There is no penalty for choosing $n_0$ too large. So guess big and see what happens. If you cannot show the inequality above for every $n$ greater than your guess of $n_0,$ guess bigger.
Once you have a good guess you can start writing your proof.
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