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How to Find a Limit at Infinity with My Friend, Bill

How to Find a Limit at Infinity with My Friend, Bill – Why A Billion is a Beautiful Approximation for Infinity

July 14, 2026 Concepts Explained Comments (0)

TLDR

Front note:

My tutoring usually involves lots of questions to help students to think about the problem and figure as much out as possible for themselves, and an entirely written explanation is somewhat different from a verbal presentation where I’m drawing and referencing equations/pictures in real-time. Usually, when I’m typing, I’m busy writing stories or blogging about writing stories, so a lot more of that kind of humor bled into this than I’d usually use. Please feel free to give feedback on whether the humor is too much or if there are places where I could be clearer. I know I can always improve. But without further ado, here is my brief explanation on how to find a limit at infinity with my good friend Bill, or in other words, why a billion is a beautiful approximation for infinity:

The End of Infinity?

Have you ever wondered what would happen at the very end of something? Like a book series, or a TV show, or an election, or a relationship, or even, perhaps a graph? That last one isn’t perhaps quite as likely, though if you’ve been in math very long, you’ll probably find that your teacher wonders for you. What happens as this graph just keeps going… forever? What if there is no end?

Or what if we take some expression,

3x4−74x8−3x7\frac{3x^4-7}{\sqrt{4x^8-3x^7}}

for example, and just feel like plugging in infinity?

(Don’t worry if you suffer from a curiosity about math. I hear that if you scratch that itch long enough, people start throwing lots of money at you, as long as you can combine that math love with something useful, like engineering, or science, or programming, or business, or becoming a professor or something. And even if they don’t, it’s still fun to learn new things!)

The Problem with Infinity

Well, unfortunately, the mathematical purists are why we can’t have nice things. They’re the ones who will insist that NO, we CANNOT just plug in infinity wherever we feel like. That would be HERESY!!!! After all, infinity is a concept of going on forever, not a number, and we can only plug numbers into expressions.**** (Well…. technically, that’s not quite true, but please ignore the man behind the curtain, still busy stuffing more straw into the other guy’s argument while grossly oversimplifying it)

Anyway, the problem with working with infinity directly is that it’s weird and doesn’t behave like a regular number. After all

∞+5=∞\infty+5=\infty

∞−73,000,000=∞\infty-73,000,000=\infty

4×∞=∞4\times\infty=\infty

∞÷7,000,000=∞\infty\div7,000,000=\infty

∞=∞\sqrt\infty=\infty

∞4000=∞\infty^{4000}=\infty

So in normal words, we can add, subtract, multiply, and divide, square root, and take infinity to basically any power we want, and it just stays infinity. That is weird! It almost acts like acid, making the rest of the expression disappear.

Take a look:

4+∞15÷7+15∞+17=∞4+\infty^{15}\div7+15\infty+17=\infty

We started with a long, complicated expression, but it all boils down to just being infinity. The other parts just disappear.

And that’s not even the worst part! Turns out that if you start messing with infinity some more, we can get “indeterminate expressions,” (sometimes called indeterminant expressions) which is math lingo for “I don’t know, this could literally be anything, so until I know more, I’ll just call it IDK,” except mathematicians like to be fancy, so they call it “indeterminate” instead of IDK.

Here are some examples:

∞−∞\infty-\infty

∞÷∞\infty\div\infty

∞0\infty^0

0∞0^\infty

0×∞0\times\infty

Basically, dealing with infinity directly causes problems because either infinity makes the entire rest of the expression disappear, or it turns into a massive IDK, or the mathematicians arrest us for handling radioactive math concepts without a license, or even gloves.

How to Find a Limit at Infinity

Now, the most common method for safely handling infinity is through the use of limits. I’ll discuss limits more thoroughly in another post, but the basic idea is that you’re looking at where a function or sequence is headed, where it’s going to. You don’t actually have to get there (though you can). This allows us to put a barrier between us and infinity, kind of like gloves. We’re not plugging in infinity directly; we’re just plugging in a variable that is approaching infinity. This is the “Mathematician Approved” method for dealing with infinity.

Unfortunately, there are times when even limits struggle with infinity, such as in the case of those “indeterminate” expressions I mentioned earlier. Yes, there are workarounds, but they get a bit complicated.

*Sigh*

Why A billion is a beautiful Approximation for Infinity

This is where my good friend Bill (short for Billion) helps out. It turns out, a billion is a beautiful approximation for infinity without all of the radioactive, blow-everything-up parts of it. After all, a billion (1,000,000,000) is HUGE! (Just like infinity) At the same time, 3 Billion is different from 2 Billion, so you can actually find a value for 3 Billion – 2 Billion, which is just 1 Billion (though in the future, I’ll just put “Bill” in the mathematical expressions instead of the full “Billion”). What would have been indeterminate if I’d plugged in infinity (∞ – ∞) becomes simple when I plug in Bill instead.

Let me show you…

How to Find a Limit at Infinity with Our Good Friend Bill

Now, let’s take another look at our random expression from earlier where we felt that urge to plug in infinity.

3x4−74x8−3x7\frac{3x^4-7}{\sqrt{4x^8-3x^7}}

Yep, it’s still a little complicated. As mentioned, the official way to handle it is to take its limit at infinity, which in math notation looks like

limx→∞⁡3x4−74x8−3x7\lim_{x \to \infty}\frac{3x^4-7}{\sqrt{4x^8-3x^7}}

But then we look at what we’d get as x approaches ∞, and we’re back to

3∞4−74∞8−3∞7\frac{3\infty^4-7}{\sqrt{4\infty^8-3\infty^7}}

We’ve got (∞ – ∞) inside the square root, and potentially ∞ ÷ ∞ in the form of a fraction, if the ∞ – ∞ turns out to equal ∞. Oh, no! More indeterminates!

But then Bill comes to save the day! Plug him in instead and we get

3Bill4−74Bill8−3Bill7\frac{3Bill^4-7}{\sqrt{4Bill^8-3Bill^7}}

Yes, that looks similar to what we had before, but watch this!

Look at the top of our fraction (the numerator, for our vocab specialists).

3Bill4−73Bill^4-7

A Billion is only an approximation for infinity anyway, so why don’t we approximate some more? If a Billion is huge, then a billion to the fourth power (Bill * Bill * Bill * Bill) is simply ginormous. And since it is so colossal, subtracting 7 from it doesn’t really do much. Tripling it is still going to make it significantly bigger, but subtracting a dinky little 7? Not so much. The 7 makes so little difference, it might as well not be there at all, allowing it to approximate the top as

3Bill43Bill^4

Now for the bottom (the denominator)!

We can use a similar logic to simplify the bottom, with a twist.

4Bill⁸ and 3Bill⁷ are both huge, but notice how Bill⁸ is the same as Bill ✕ Bill⁷.

That means that 4Bill⁸ is roughly A BILLION TIMES BIGGER THAN 3Bill⁷.

So even though they’re both huge, 4Bill⁸ is WAY bigger, so much bigger, in fact, that we can safely say that

4Bill8−3Bill74Bill^8-3Bill^7

is basically the same as

4Bill84Bill^8

(It’s worth mentioning that any Billion to a higher power is so much bigger than a Billion to a lower power that when adding/subtracting them, the Billion to the lower power can be ignored.)

Using those 2 approximations, we get

3Bill4−74Bill8−3Bill7≈3Bill44Bill8\frac{3Bill^4-7}{\sqrt{4Bill^8-3Bill^7}}\approx\frac{3Bill^4}{\sqrt{4Bill^8}}

(The ≈ symbol can read as “is approximately” or “is roughly”)

And now that we’ve got something simpler, we can continue to simplify it even more! After all,

3Bill44Bill8=3Bill42Bill4=32{\frac{3Bill^4}{\sqrt{4Bill^8}}=\frac{3Bill^4}{2Bill^4}}=\frac{3}{2}

Ta Da!

Isn’t Bill awesome?

Bill helped us to find the limit at infinity by approximating infinity in a way that let us preserve our natural intuition about numbers and their relative sizes. Go Bill!

TLDR

Plugging in infinity causes big problems, since many aspects of infinity break our intuition. Limits are the official way to deal with infinity, but plugging in a Billion and using the fact that a Billion minus 7 is basically just a Billion allows us to simplify things a lot. Similarly, since a Billion squared (which I’ll write as Bill²) is a billion billions (Bill² = Bill ✕ Bill), then even a number as big as 95000Bill is so much smaller than Bill² that

Bill2−95000Bill≈Bill2Bill^2 – 95000Bill\approx Bill^2

Further, any Billion to a higher power is so much bigger than a Billion to a lower power that when adding/subtracting, the Billion to the lower power can be ignored.

We can use this method of simplification to simplify all the additions/subtractions in our expressions as much as possible before applying any other mathematical operations like multiplication, division, square roots, etc.

Example:

limx→∞⁡3x5−74x8−3x7≈3Bill5−74Bill8−3Bill7≈3Bill54Bill8=3Bill52Bill4=32Bill≈32∞=∞\lim_{x \to \infty}\frac{3x^5-7}{\sqrt{4x^8-3x^7}}\approx\frac{3Bill^5-7}{\sqrt{4Bill^8-3Bill^7}}\approx\frac{3Bill^5}{\sqrt{4Bill^8}}=\frac{3Bill^5}{2Bill^4}=\frac{3}{2}Bill\approx\frac{3}{2}\infty=\infty

This method can help you to simplify a wide variety of limits at infinity. Isn’t a Billion a beautiful approximation for infinity?

Examples of Finding a Limit at Infinity with Our Good Friend Bill

Algebra I or Algebra II

End Behavior of a polynomial:

If

f(x)=2x3−13x2−7x−1827f(x)=2x^3-13x^2-7x-1827

then when x → ∞

2x3−13x2−7x−1827≈2Bill3−13Bill2−7Bill−18272x^3-13x^2-7x-1827\approx2Bill^3-13Bill^2-7Bill-1827

≈2Bill3−13Bill2−7Bill≈2Bill3−13Bill2\approx2Bill^3-13Bill^2-7Bill\approx2Bill^3-13Bill^2

≈2Bill3≈2∞3=2∞=∞\approx2Bill^3\approx2\infin^3=2\infin=\infin

Precalculus

End Behavior of a Rational Function:

If

f(x)=3x5−4x3+2x6+3x4−2x2−87f(x)=\frac{3x^5-4x^3+2}{x^6+3x^4-2x^2-87}

then when x → -∞,

3(−Bill)5−4(−Bill)3+2(−Bill)6+3(−Bill)4−2(−Bill)2−87≈−3Bill5+4Bill3Bill6+3Bill4−2Bill2≈−3Bill5Bill6+3Bill4≈−3Bill5Bill6≈−3Bill≈0\frac{3(-Bill)^5-4(-Bill)^3+2}{(-Bill)^6+3(-Bill)^4-2(-Bill)^2-87}\approx\frac{-3Bill^5+4Bill^3}{Bill^6+3Bill^4-2Bill^2}\approx\frac{-3Bill^5}{Bill^6+3Bill^4}\approx\frac{-3Bill^5}{Bill^6}\approx\frac{-3}{Bill}\approx0

Precalculus, AP Calculus AB, or Calculus 1

Find a limit at infinity:

limx→∞⁡7(x6+2x4−7x+1)2/3|4x4−3x3+189x2−576|≈7(Bill6+2Bill4−7Bill+1)2/3|4Bill4−3Bill3+189Bill2−576|≈7(Bill6+2Bill4−7Bill)2/3|4Bill4−3Bill3+189Bill2|\lim_{x \to \infty}\frac{7(x^6+2x^4-7x+1)^{2/3}}{|4x^4-3x^3+189x^2-576|}\approx\frac{7(Bill^6+2Bill^4-7Bill+1)^{2/3}}{|4Bill^4-3Bill^3+189Bill^2-576|}\approx\frac{7(Bill^6+2Bill^4-7Bill)^{2/3}}{|4Bill^4-3Bill^3+189Bill^2|}

≈7(Bill6+2Bill4)2/3|4Bill4−3Bill3|≈7(Bill6)2/3|4Bill4|=7Bill44Bill4=74\approx\frac{7(Bill^6+2Bill^4)^{2/3}}{|4Bill^4-3Bill^3|}\approx\frac{7(Bill^6)^{2/3}}{|4Bill^4|}=\frac{7Bill^4}{4Bill^4}=\frac{7}{4}

AP Calculus BC or Calculus 2

Determining whether a series will converge or diverge:

Even though you’ll have to use another method to actually prove it, at least it can give you an idea of which functions might work best for a direct comparison or limit comparison test:

∑n=0∞5n−75n6−4n4+73≈∑n=0∞5Bill−75Bill6−4Bill4+73≈∑n=0∞5Bill5Bill63≈∑n=0∞5Bill53Bill2≈∑n=0∞553Bill≈553∑n=0∞1n\sum_{n=0}^{\infin} \frac{5n-7}{\sqrt[3]{5n^6-4n^4+7}}\approx\sum_{n=0}^{\infin} \frac{5Bill-7}{\sqrt[3]{5Bill^6-4Bill^4+7}}\approx\sum_{n=0}^{\infin} \frac{5Bill}{\sqrt[3]{5Bill^6}}\approx\sum_{n=0}^{\infin} \frac{5Bill}{\sqrt[3]{5}Bill^2}\approx\sum_{n=0}^{\infin} \frac{5}{\sqrt[3]{5}Bill}\approx\frac{5}{\sqrt[3]{5}}\sum_{n=0}^{\infin} \frac{1}{n}

which diverges. Therefore, since the original series is approximately

553∑n=0∞1n\frac{5}{\sqrt[3]{5}}\sum_{n=0}^{\infin} \frac{1}{n}

which diverges,

∑n=0∞5n−75n6−4n4+73\sum_{n=0}^{\infin} \frac{5n-7}{\sqrt[3]{5n^6-4n^4+7}}

almost certainly diverges, too, and we can prove it using the direct comparison test with

∑n=0∞1n\sum_{n=0}^{\infin} \frac{1}{n}

Proof: Since either decreasing the top (numerator) or increasing the bottom (denominator) of a fraction decreases the total fraction (both ¼ and ⅗ are less than ¾),

∑n=0∞5n−75n6−4n4+73≥∑n=0∞5n−n5n6−4n4+73\sum_{n=0}^{\infin} \frac{5n-7}{\sqrt[3]{5n^6-4n^4+7}}\geq\sum_{n=0}^{\infin} \frac{5n-n}{\sqrt[3]{5n^6-4n^4+7}}

(when n ≥ 7), and

∑n=0∞5n−n5n6−4n4+73≥∑n=0∞5n−n5n6−4n4+n43=∑n=0∞4n5n6−3n43>∑n=0∞4n5n63=453∑n=0∞nn2=453∑n=0∞1n\sum_{n=0}^{\infin} \frac{5n-n}{\sqrt[3]{5n^6-4n^4+7}}\geq\sum_{n=0}^{\infin} \frac{5n-n}{\sqrt[3]{5n^6-4n^4+n^4}}=\sum_{n=0}^{\infin} \frac{4n}{\sqrt[3]{5n^6-3n^4}}>\sum_{n=0}^{\infin} \frac{4n}{\sqrt[3]{5n^6}}=\frac{4}{\sqrt[3]{5}}\sum_{n=0}^{\infin} \frac{n}{n^2}=\frac{4}{\sqrt[3]{5}}\sum_{n=0}^{\infin} \frac{1}{n}

Therefore,

∑n=0∞5n−n5n6−4n4+73>453∑n=0∞1n\sum_{n=0}^{\infin} \frac{5n-n}{\sqrt[3]{5n^6-4n^4+7}}>\frac{4}{\sqrt[3]{5}}\sum_{n=0}^{\infin} \frac{1}{n}

when n ≥ 7, and since

∑n=0∞1n\sum_{n=0}^{\infin} \frac{1}{n}

diverges, by the direct comparison test,

∑n=0∞5n−n5n6−4n4+73\sum_{n=0}^{\infin} \frac{5n-n}{\sqrt[3]{5n^6-4n^4+7}}

also diverges.

Conclusion: Now you know how to find a Limit at Infinity!

So, the next time you’re stuck facing the infinite hordes and wondering how things will turn out, or any of the infinite ways that infinity can be frustrating, just remember your old friend Bill and appreciate the fact that a billion is a beautiful approximation of infinity. It might not always be the “correct” way to find a limit at infinity (as we saw in the convergence/divergence example), but at least it will be an easy, fast, and convenient option to know if you’re on the right track.

And when you can handle infinity, the possibilities are limitless! 😉

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