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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 $0$.

enter image description here

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.

enter image description here

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.

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    $\begingroup$ Basically it means the percentage difference between the vertical distances between points on the two graphs approaches $0.$ $\endgroup$ Dec 23 '20 at 20:16
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    $\begingroup$ 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 $f/g\to 1$. $\endgroup$ Dec 24 '20 at 3:32
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    $\begingroup$ 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. $\endgroup$
    – Joe
    Dec 24 '20 at 16:57
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$x^2+x$ 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:

Semi-log plot

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 $y=(x^2+x)/x^2$: f/g plot

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    $\begingroup$ 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 :) $\endgroup$
    – chaad
    Dec 23 '20 at 18:29
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    $\begingroup$ @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... $\endgroup$
    – Joe
    Dec 23 '20 at 18:38
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    $\begingroup$ @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. $\endgroup$
    – Joe
    Dec 23 '20 at 18:39
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    $\begingroup$ @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. $\endgroup$
    – Joe
    Dec 23 '20 at 18:49
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    $\begingroup$ @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. $\endgroup$
    – Joe
    Dec 23 '20 at 19:16

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