Basic question on limits I have one basic doubt. I want to know $\displaystyle \lim_{n \to \infty}  f(n)$ in each of the three cases:


*

*$\displaystyle \lim_{n \to \infty} \frac{f(n)}{n}=0$.

*$\displaystyle \lim_{n \to \infty}  \frac{f(n)}{n}=c$ where $c>0$.

*$\displaystyle \lim_{n \to \infty}  \frac{f(n)}{n}=c$ where $c<0$.
 A: Conceptually:
$$\lim_{n \rightarrow \infty}f(n) = \lim_{n \rightarrow \infty}\left(\frac{f(n)}{n}\right)\cdot n = \left(\lim_{n \rightarrow \infty} \frac{f(n)}{n}\right) \cdot \left(\lim_{n \rightarrow \infty} n\right) = \left(\lim_{n \rightarrow \infty} \frac{f(n)}{n}\right) \cdot \infty$$
In the first case, we get $0 \cdot \infty$, which is indeterminate (in other words, there is not enough information to determine $\lim_{n \rightarrow \infty}f(n)$), in the second case we get $c \cdot \infty = \infty$, since $c > 0$, and in the third case we get $c \cdot \infty = -\infty$ since $c < 0$. 
As you can probably tell, this is not quite rigorous, but it can be a good way to start thinking about these kinds of problems.
A: In the first case, you can have literally any behavior: for any real number $a$, we can let $f$ be the constant function $f(x)=a$, in which case we have
$$\lim_{n\to\infty}f(n)=\lim_{n\to\infty}a=a,\qquad \lim_{n\to\infty}\frac{f(n)}{n}=\lim_{n\to\infty}\frac{a}{n}=0,$$
we can let $f(x)=\begin{cases}\sqrt{x} & \text{ if }x\geq 0,\\ 0 & \text{otherwise}\end{cases}$, in which case we have
$$\lim_{n\to\infty}f(n)=\lim_{n\to\infty}\sqrt{n}=\infty,\qquad \lim_{n\to\infty}\frac{f(n)}{n}=\lim_{n\to\infty}\frac{1}{\sqrt{n}}=0,$$
and we can let $f(x)=\begin{cases}-\sqrt{x} & \text{ if }x\geq 0,\\ 0 & \text{otherwise}\end{cases}$, in which case we have
$$\lim_{n\to\infty}f(n)=\lim_{n\to\infty}-\sqrt{n}=-\infty,\qquad \lim_{n\to\infty}\frac{f(n)}{n}=\lim_{n\to\infty}-\frac{1}{\sqrt{n}}=0.$$
In the second case, if $\lim_{n\to\infty}\frac{f(n)}{n}=c$ and $c>0$, then for any $\epsilon>0$, there is an $N$ such that $|\frac{f(n)}{n}-c|<\epsilon$ for all $n>N$; for example, letting $\epsilon=\frac{c}{2}$, we have that $|\frac{f(n)}{n}-c|<\frac{c}{2}$ for all $n>N$ for some $N$. Thus
$$\frac{f(n)}{n}>\frac{c}{2}$$
for all $n>N$, which implies that $f(n)>n\frac{c}{2}$ for all $n>N$, and hence $\lim_{n\to\infty}f(n)=\infty$.
A similar analysis of the third case will demonstrate that we have to have $\lim_{n\to\infty}f(n)=-\infty$.
