Use the definition of a limit to prove that the limit is equal to zero? 
All I can think of to start is to state that:
$$|n-∞| < \delta \Rightarrow |(c/n^2)-0| < \epsilon$$
But I don't know where to go from there
 A: You want to prove that $(C/n^2)\to 0$ as $n\to\infty$.
You are required to use the definition that: $$
\lim_{n\to\infty}a_n = l \quad\iff\quad \forall \epsilon>0:\exists N:\forall n>N: \big(\lvert a_n-l\lvert < \epsilon \big) 
$$
Since $a_n=C/n^2$ and $l=0$ you must show that $$
\lim_{n\to\infty}\frac C{n^2} = 0 \quad\iff\quad \forall \epsilon>0:\exists N:\forall n>N: \big(\Big\lvert \frac C{n^2}\color{silver}{-0}\Big\rvert < \epsilon \big)$$
A: Look at the definition they have given. You want $\ell=0$ in this case. Given any $\epsilon>0$, you have to find some $N$ such that for any $n>N,\; \left|\frac C{n^2} - \ell\right|<\epsilon$. See if you can work from there.
A: Danielle,
First, the presence of the constant $C$ is immaterial to the limit. You can factor is out and reduce the question to $C \lim_{n \to \infty} \dfrac{1}{n^2} = 0 \implies \lim_{n \to \infty} \dfrac{1}{n^2} = 0$.Then your $|a_n - \ell|$ is fairly straightforward, as $\ell = 0$ means that $|a_n - 0| = a_n$. Choose such an $\epsilon = \dfrac{1}{N} > 0$. Now, just solve for $n$; so that for sufficiently large $n$ we have $| \dfrac{1}{n^2} | < \dfrac{1}{N^2} < \epsilon^2 < \epsilon \space \forall n > N$.
A: $lim_{n \to \infty}$ $C/n^2=0$ <=> $C*lim_{n\to\infty}1/n^2=0$ 
<=> $lim_{n\to\infty} 1/n^2=0$
Given any $\epsilon \gt 0$, there is some N such that for any $n \gt N$,∣∣$1/n^2−0$∣∣$\lt\epsilon$.
$∣∣1/n^2−0∣∣\lt\epsilon 
<=> ∣∣1/n^2∣∣\lt\epsilon
<=> 1/n^2\lt\epsilon 
<=> n^2 > 1/\epsilon 
<=> n \gt \sqrt{1/\epsilon}
$
Let $N = \sqrt{1/\epsilon}$. Then whenever $n\gt N$, $∣∣1/n^2−0∣∣\lt\epsilon$ 
