Math

# The Physics Philes, lesson 26: Euler’s Detour

In which limits is finished, special theorems are listed, and a slight mathematical detour is taken.

It’s time! It’s time, ladies and gentleman, for me to finally – FINALLY – finish up the two chapters on limits. Two chapters that I wanted to take two posts has taken four, but there you have it. It’s time to finish up this topic.

Last week I attempted to explain limits and vertical asymptotes. Today we’ll look at limits and horizontal asymptotes and some special limit theorems. But first! Horizontal asymptotes.

A horizontal asymptote pretty much just looks like a vertical asymptote turned on its side. It’s the height the function tries to reach, but cannot, as the function’s x-value gets infinitely more positive or negative. In other words, instead of going on forever up and/or down, it goes on forever right and/or left. We call horizontal limits “limits at infinity” because it’s not approaching a fixed number like typical limits. The limit still exists because the function intends to reach the limiting height.

Evaluating horizontal limits is actually pretty easy, even though the methods we’ve learned so far for evaluating limits (substitution, factoring, and conjugating) don’t work. All we have to do is compare the highest exponents in the numerator and denominator. There are three rules, but first you need to know a couple of definitions. A degree of a polynomial is the value of the largest exponent. The leading coefficient of a polynomial is the coefficient of the term with the largest exponent.

OK. The three rules:

1. If the degree of the numerator is higher than the degree of the denominator, then the limit of the function is equal to infinity or negative infinity.
2. If the degree of the denominator is higher than the degree of the numerator, then the limit of the function is equal to zero.
3. If the degrees are equal, then the limit is equal to the leading coefficient of the numerator divided by the leading coefficient of the denominator.

These three rules only applies to limits at infinity. Don’t try this with other types of limits. People could get hurt. Let’s see how this works with a couple of examples.

Evaluate this limit: Holy cow! So complicated! Nope! Not really. All we have to do is follow the rules. The degree in both the numerator and denominator is 3, so the first two rules are no help. So we need to use the third rule. The leading coefficient of the numerator is 5 and the leading coefficient of the denominator is 8. If we follow the third rule, that means the answer is: Wow. That’s…easy. Let’s do one more. Evaluate: The degree of the numerator is 2 and the degree of the denominator is 3. That means that, following rule 2, the answer is Still super easy, right? I think it’s time to move on. Time for some special limit theorems.

There are, evidently, a few limit theorems that don’t show up very often but that, if you want to learn calculus, you need to know anyway. I’ll just list them. This is only true when you approach zero and α can be any quantity. Like the previous theorem, this is only true when approaching zero. It’s sometimes written as 1 – cos α/α, but don’t worry about it. The answer is still zero. This theorem basically says that if any real number is divided by x and we let x get infinitely large the result will be zero. This makes sense. If we divide 4 by a bajillion, the answer is going to so small that it might as well be zero. This looks complicated, but it just says that one plus an extremely small number, when raised to an extremely high power, is exactly equal to Euler’s number.

Wait. One. Second. Euler’s number? Way to have no explanation, book! We need a quick detour. What the heck is Euler’s number?

Euler’s number (e) is one of those mathematical constants and it’s approximately equal to 2.71828. Euler’s number is an irrational number and it’s pretty important for calculus because it allows us to do differential and integral calculus with exponential functions and logarithms.

I don’t know what any of that means. You smarty smarts can try to explain it in the comments if you want, but I’m just hoping that it will start to make sense as my calculus knowledge increases. Anyhoo, detour over! Let’s work a couple of examples. Evaluate: Alrighty, this looks a little like the first theorem. But the first theorem only works when the value inside the sine matches the denominator. So we need 3x in the denominator.  We can evaluate the limits of the factors separately and multiply the results together to get the final answer. See? The first expression is equal to 1 via the first theorem. Which is why we multiple 3 by 1 to get an answer of 3. Get it?

Let’s do one more. Evaluate: Now we just plug in those values to solve:

Whew! Finally! Done with limits. I’m not sure what I’ll move on to next week. Either continuity or something physic-y. Only time will tell.

Featured image credit: Mark Turnaukas