# Visual intuition for the definition of "asymptotically equivalent"

I'm trying to intuitively grasp the following definition:

The real-valued functions $$f$$ and $$g$$ are asymptotically equivalent as $$x \to \infty$$ if $$\lim_{x \to \infty} \dfrac{f(x)}{g(x)}=1.$$ We write this as $$f \sim g$$.

My question is: how do we visually interpret this in terms of the graphs of $$f$$ and $$g$$? Does this mean that the graphs of $$f$$ and $$g$$ get closer to each other as $$x$$ gets larger and larger?

My only intuition for this comes from the following example: we know that $$\sin x \sim x$$ as $$x \to 0$$ (since $$\lim_{x \to 0} (\sin x)/x = 1$$). And as we can see below, the graphs of $$\sin x$$ (the green line) and $$x$$ (the black line) get closer and closer as $$x$$ goes to $$0$$. But this intuition does not seem to hold for functions asymptotically equivalent at $$\infty$$. I graphed $$x^2 + x$$ (black line) and $$x^2$$ (green line) and their graphs do not appear to be getting closer at all! In fact, it looks like there's a "gap" between the two graphs. This leads me to believe that I'm not interpreting "asymptotically equivalent" in the right way. I've come across the idea that $$f \sim g$$ means that $$f$$ and $$g$$ have the "same rate of growth", but that feels very unintuitive for me. Is there are a way to see that in the graphs?

Any guidance would be greatly appreciated! Thanks.

• Basically it means the percentage difference between the vertical distances between points on the two graphs approaches $0.$ Dec 23 '20 at 20:16
• Draw the graph of $f/g$ and it will get closer and closer to $1$ as you move towards right in the graph. You can't ensure $f$ close to $g$. For that you need $f-g\to 0$ and not $f/g\to 1$. Dec 24 '20 at 3:32
• By the way, I think this is an excellent question since asymptotic equivalence is a very powerful tool in maths. It can be used to give intuitive justifications to a number of theorems, and is also not too hard to turn these justifications into rigorous arguments.
– Joe
Dec 24 '20 at 16:57

$$x^2+x$$ and $$x^2$$ are asymptotically equivalent, since $$\lim_{x \to \infty}\frac{x^2+x}{x^2}=\lim_{x \to \infty}1+\frac{1}{x}=1 \, .$$ So it is probable that you simply didn't choose $$x$$ values that were large enough to make your intuitions work. For example, $$x=100$$ gives $$\frac{100^2+100}{100^2}=\frac{10100}{10000}=1.01 \, ,$$ and you can see that they are very close to each other in relative terms. Notice the use of the word relative here. The example you gave helps illustrate what I mean. Let $$f(x)=x^2+x$$ and $$g(x)=x^2$$. $$f(x)-g(x) \to \infty$$ as $$x \to \infty$$. This means that in absolute terms, the two functions are growing apart. However, the ratio between them—what you need to multiply $$x^2$$ by to get $$x^2+x$$—is approaching $$1$$. This is what the notion of 'asymptotically equivalent' is trying to capture.

To visualise this, it would be better to use a logarithmic scale: Look at $$x=10$$, for instance. $$x^2=100$$, whereas $$x^2+x=110$$. This discrepancy looks small when we use a logarithmic y-axis, since the gap of $$10$$ is small relative to how large the functions are. Or, as Paramanand Singh has suggested, we could plot $$y=(x^2+x)/x^2$$: • My follow-up question would be: how do we interpret relative difference using the graphs of $f$ and $g$? From the picture you posted, it looks like the graphs $f$ and $g$ are "hugging" each other; but since $f(x)-g(x) \to \infty$ as $x \to \infty$ so wouldn't the graphs eventually grow arbitrarily far apart? That's why I'm struggling to visually interpret the notion of "relative difference". Any tips would be appreciated :) Dec 23 '20 at 18:29
• @chaad Imagine plotting the two graphs on a $1:1$ scale. This would be very awkward, especially for large values of $x$. It will probably get to a point where you have to scroll on your computer to go from $g$ to $f$. The amount of scrolling you have to do is proportional to the distance between $f(x)$ and $g(x)$. This is what 'absolute difference' might mean intuitively. Then, imagine using a more reasonable scale. If both $f(x)$ and $g(x)$ are very big, then it seems silly to worry about the differences between them (more precisely, the distances between them are tiny relative to...
– Joe
Dec 23 '20 at 18:38
• @chaad ...the size of $f(x)$ and $g(x)$ themselves). This is what 'relative difference' is trying to capture. In a sense, $f(x)$ and $g(x)$ should be hugging. This perhaps shows the deficiencies in linear scales—they don't show us the big picture. They exaggerate discrepancies, which, in a relative sense, are tiny.
– Joe
Dec 23 '20 at 18:39
• @chaad NB If you look closely, then you can still see $f(x)$ and $g(x)$ moving apart in the graph. It's just that we don't consider this difference to be very important.
– Joe
Dec 23 '20 at 18:49
• @chaad Here is a plot of $\log(x^2+x)$ versus $\log(x^2)$. (I used the base-$10$ logarithm here, but any is fine.) Because we are using a more 'reasonable' scale, the two graphs are actually coming together, not going apart. Logarithmic scales are in many ways preferable, since they don't emphasise small discrepancies like linear scales do. I hope that answers your question.
– Joe
Dec 23 '20 at 19:16