# The Physics Philes, lesson 24: Know Your Limits

*In which decimal places are avoided, conjugates are explained, and many more limit problems are completed.*

Last week I tried to give an introductory overview of limits, what they are, and how they work. You may have thought (as I did) that evaluating limits sounds like a nightmare. I mean, you could just plug a number into a function that is extremely close to the number we are approaching, but that can get pretty tedious. There must be another way.

Luckily there is another way. In fact, there are three methods we can use to evaluate limits that don’t involve infinite decimal places.

The first of these methods is the substitution method. This method is pretty easy. All you have to do is plug in the x-value you are approaching into the function. Let’s evaluate a limit in this way to see how it works.

Just replace 4 for x and solve:

So according to this method, as x approaches 4, the function equals 14:

The second method for evaluating limits is the factoring method. (Man, that factoring stuff keeps cropping up, doesn’t it?) Let me explain it by way of an example.

We can cancel out (x + 3), so we’re left with:

Then we use the substitution to finish it off.

Again, it’s super easy, right? Soak it in, because the final method is a little bit more complicated. But just a little bit.

The final method for evaluating limits is the conjugate method. This method is useful for limits that contain radicals. To understand this method, we need to know what a conjugate is.

Essentially, the conjugate of a binomial expression is the same expression with opposite middle sign. For example, the conjugate of √x – 5 is √x + 5. The products of two conjugates containing radicals will have no radicals in it. Like this:

Like magic! But it’s not magic. It’s math.

Anyhoo, you use the conjugate method when you are working with a limit problem that contains radicals for which substitution doesn’t work. Let’s work on an example.

This equation has a radical and a number subtracted from it. This is a sign that you should use the conjugate method. So let’s get to it! All you have to do is multiply the numerator and the denominator by the conjugate of the radical expression.

Multiply the numerators first.

Now we have an equation that looks like this:

See why we didn’t actually multiply together the denominator? Now we can cancel out x – 5 to get this:

Look! Now we can use the substitution method to finish it off! The x-value we are approaching is 5, so:

That means that the solution for this limit problem is:

Ta da! Not too complicated. It just has a few more steps.

But what if none of these methods works? Are we just stuck? Nope. In that case, you just need to plug in a number that is extremely close to the number being evaluated. Hopefully those types of problems don’t crop up very often.

Now you know how to evaluate limits! Not too bad, if I say so myself. I don’t know what I was so scared of. But we’re not done. Next week we’ll check out some special limits theorems and the relationship between limits and infinity. You’re not going to want to miss that.

*Featured image credit: wintercool612*

## 4 Comments

### Leave a Reply

You must be logged in to post a comment.

Excellent and fun, as usual.

Aw, shucks. I try.

Hi Mindy,

Good job explaining how limits work.

It seems like you’re getting ready to dive into calculus. I look forward to it.

For now, I’d like to point out another small mistake:

sqrt(x*x+11*11) is not equal to x + 11.

Basically, change the formula after “Multiply the numerators first.” to:

(sqrt(x+11))^2 + …

where

sqrt means square root and

^2 means squared.

I know that it’s not a big deal in the grand scheme of things, but… 😛

There is no mistake too small! I’m pretty open about the fact that I don’t know what I’m doing Thanks for the correction!