$h \in O(h^2)$ as $h \rightarrow 0$?

Question

My question is : $h \in O(h^2)$ as $h \rightarrow 0$ ?

I am only used to use this notation for sequences when $n \rightarrow \infty$. In this case I know that $n \in O(n^2)$.

Can someone confirm it?

Answer 1

Usually it is easier to look at the limit definition of big-$\mathcal O$:

$g\in\mathcal O(f)$ at $a\iff \limsup_{h\to a}\left|\frac{g(h)}{f(h)}\right|<\infty$

You can think on big-$\mathcal O$ like: If you get close to $a$, does $f$ grow faster than $g$?

You can see that forall $|h|<1$ there is $h^2<h$, moreover when $h\to0$ there is $h^2\ll h$, hence $h\notin\mathcal O(h^2)$.

In the limit definition it is immediate:$$\limsup_{h\to0}\left|\frac{h}{h^2}\right|=\lim_{h\to0}\left|\frac{h}{h^2}\right|=\lim_{h\to0}\left|\frac{1}{h}\right|=\lim_{h\to0}\frac{1}{|h|}=\infty\require{cancel}\cancel<\infty$$

Answer 2

HINT

Simply note that

$$\lim_{h\to 0} \left|\frac h {h^2}\right|=\lim_{h\to 0} \left|\frac 1 {h}\right|=\infty$$

Answer 3

In a situation like this, I recommend plugging in some numbers. Here's a table of values: $$ \begin{array}{c|lll} h & .1 & .01 & .001 & .0001 & .00001 & .000001 & .0000001 \\ \hline h^2 & .01 & .0001 & .000001 & .00000001 &.0000000001 & .000000000001 & .00000000000001 \end{array} $$ So, do you still believe that $h < ch^2$ for some $c$ as $h \to 0$?

Answer 4

Let's apply a change of variables where $g=\frac{1}{h}$. Note that as $h\to 0$ then $g \to \infty$.

Your question asks whether $\frac{1}{g} = \frac{1}{g^2}$ as $g \to \infty$

Clearly, as $g \to \infty$, $$\frac{1}{g} \not \in O(\frac{1}{g^2})$$ Since

$$\lim_{g \to \infty} \left | \frac{\frac{1}{g}}{\frac{1}{g^2}} \right | = \lim_{g \to \infty} \left | g \right| \to \infty$$

Therefore, for all $g>g_0 >0$, $\not \exists M > 0$ such that

$$\left | \frac{1}{g} \right | < M \cdot \left | \frac{1}{g^2} \right |$$

Which is equivalent to for all $0<h < h_0$ $\not \exists M > 0$ such that $$\left | h \right | < M \cdot \left | h^2 \right |$$

Note: Other answers are more direct, this is a roundabout way to get to the same result.