Limits by Piecewise Function and the Intuition Behind the Limit: An Introduction to Calculus

In our previous article, we introduced the concept of limits by rationalization, and worked through why it works.

To combine some aspects of limits we’ve already learned, let’s do some more problems, now using piecewise functions, and discuss the one piece of intuition about limits I have purposely been leaving out.

“Limits by Piecewise Functions” is actually kind of a misnomer, as it errs more on the side of an application of limits than it is a way to solve limits.

Nevertheless, it is a helpful tool to algebraically represent complex limits that cannot be defined with a single function.

Take a look at the following piecewise function:

At first glance, this looks a little complicated. After all, what are inequalities doing in a function?

As it turns out, all the notation on the right just determines where the parts of the function begin and terminate.

Put another way, f(x)=(x+2)/(x-1) when x is greater than zero but less than or equal to 4, and f(x)=sqrt(x) when x is larger than 4.

Let’s observe how the function looks graphically:

What is the limit as x→4+ (approaches 4 from the right)?

Well, we know that when x is greater than 4, f(x)=sqrt(x).

When we approach 4 from the right side, we mean that x gets infinitely close to four, but x is always still greater than 4.

Thus, the limit as x→4+ is sqrt(4), which is just equal to 2.

Remember, even though we technically plugged in the value 4, which should be defined as (4+2)/(4–1), as we approach infinitely close 4 four from the right, we can say that “infinitely close to four” is essentially the same as “four”, and thus obtain the correct answer 2.

I recognize that this is a very handwavy method of justification, but without discussing a formal definition of the limit, which generally not considered important beyond situations such as these, it is difficult to establish why this is. If you happen to know of any methods, please let me known in the comments.

However, don’t get confused. This only works if x really is getting infinitely close to four. If x is any finite distance away from 4, and x is less than four, then the function must be defined as (x+2)/(x-1).

This is also a good time to practice some intuition for why this works:

1. We can use finite distances to approximate limits.

Using the previous question as an example, when x→4+, x is getting infinitely close to four from the right.

So let’s say that x is approximately 4.0000001. This is still greater than 4, so f(x)=sqrt(x).

So, to evaluate the limit, just evaluate f(4.0000001). This is equal to sqrt(4.0000001), which is around 2.000000025. That’s basically just 2, which was the correct answer.

2. Don’t abuse 1.

You’ll notice that this method of approximation actually works for any limit. For example:

Because plugging in 2 for x doesn’t work, instead of our usual method of limits by factoring, let’s say that x is approximately 2.0001, and evaluate the limit that way.

This gives:

Which is equal to around 5.0001, which is basically just 5, which is the correct answer.

You might think that instead of learning all these methods to evaluate limits such as by factorization or rationalization, you should just evaluate limits like this. And you can. But doing so this way is slow, prone to errors, and often requires calculators.

This intuition should be just that. When you evaluate a limit, and are unsure of if it is the correct answer, this is just one way you can confirm if it is.

Think of this like the algebraic equivalent of checking if you got a limit correct by graphing it.

Going back to the original question, what about the limit as x→4- (approaches 4 from the left)?

Well, when x approaches 4 from the left, we get infinitely close to 4.

From our previously mentioned intuition, we can imagine this as x approaching values like 3.9, then 3.99, then 3.999, etc.

In all these scenarios, x is always less than four, which works out for us because when x is between 0 and 4, f(x)=(x+2)/(x-1).

In this way, we can plug in 4 for x, which gives us (4+2)/(4–1)=2.

Finally, what about the limit as x→4?

There’s actually two ways to go about solving this:

1. We know that when x=4, f(x) is still defined as (x+2)/(x-1). Thus, simply plug in 4 for x, which gives us the correct answer of 2.

2. Recall from earlier that:

We’ve already established that the limit as x→4+=2, and that the limit as x→4-=2. So, using this definition, when both one-sided limits are equal to each other, the limit is equal to either one-sided limit. Therefore, the limit as x→4 is equal to 2.

This problem was an example of limits via piecewise functions, and as you’ve noticed, used many of the different limit-solving techniques used in the past.

Try solving the problem again, and in doing so, you will further strengthen the techniques you already know.

This is the eigth installation in a series of articles attempting to explain the entirety of Calculus AB to a mainstream audience. It is designed to be easily understood, without resorting to dumbing down complex and nuanced topics. A full article list can be found on my profile.