# Intuition behind two functions being equal up to order $n$ at $a$

Two functions $f$, $g: \mathbb R \to \mathbb R$ are said to be equal up to order $n$ at $a$ if and only if $$\lim_{x \to a} \frac{f(x) - g(x)}{(x - a)^n} = 0.$$ What is the intuition behind this definition? What is the idea behind considering two functions to be "very equal" versus "not so equal"?

The current picture in my head is this: two functions are "somewhat equal" if their graphs "look like they overlap." Given $a \in \mathbb R$, two functions are equal up to a "high order" if you have to zoom in "a lot" at $a$ in order to tell the two graphs apart; they are equal up to a "low order" if you can see two distinct graphs even without zooming in at $a$? I don't really know how I got this picture; I think it's somewhat correct, but I can't see how the picture relates to the limit definition.

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Actually, if two functions agree to first order, then the more you zoom in, the less you can tell the difference between them! They have the same slope at $a$, but they bend in different ways, and you have to "zoom out" a bit to see that. (I'm assuming that by "zooming" what we mean is that we're looking at the graphs of $y = f(x)$ and $y = g(x)$ and scaling both $x$ and $y$ by the same amount, preserving the aspect ratio as if zooming into a picture.) – Rahul Jun 17 '12 at 4:21

Expanding on the insightful comment of Rahul Narain, one could say that the two functions are equal up to order $n$ if you can't tell the difference between the two graphs when you zoom in on their common point $(a,f(a))$ using a zoom factor $K$ for the $x$ variable and a factor $K^n$ for the $y$ variable, and let $K$ grow without bound. (Really, you should look at the graph of $f-g$ instead, zooming in on $(a,0)$, and try to separate it from the $x$ axis when thinking of it this way, as the graphs of $f$ and $g$ individually will appear very steep otherwise.)
Think about $h(x) := f(x)-g(x)$. If $\lim_{x \rightarrow a} \frac{h(x)}{(x-a)^n} = 0$, then that means that the Taylor series for $h(x)$ vanishes for the first $n$ terms.
Geometrically, if you draw a picture of $(x-a)^n$, you'll see that it goes to zero as $x \rightarrow a$ faster and faster (near $a$) for larger $n$. That this limit is zero means that $f(x)-g(x)$ is squeezed inside the region determined by $(x-a)^n$ (or possibly its negative, I guess).
Under the assumption, of course, that $\,h(x)\,$ is derivable n times around $\,a\,$ , something the OP didn't write. I though agree with you. – DonAntonio Jun 17 '12 at 4:11