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### Course: AP®︎/College Calculus BC > Unit 1

Lesson 6: Determining limits using algebraic properties of limits: direct substitution- Limits by direct substitution
- Limits by direct substitution
- Undefined limits by direct substitution
- Direct substitution with limits that don't exist
- Limits of trigonometric functions
- Limits of trigonometric functions
- Limits of piecewise functions
- Limits of piecewise functions
- Limits of piecewise functions: absolute value

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# Limits of piecewise functions: absolute value

This video focuses on finding the limit of |x-3|/(x-3) at x=3 by rewriting it and examining it as a piecewise function. This approach helps us understand the behavior of the function for x values greater or less than 3, revealing that the limit doesn't exist. Created by Sal Khan.

## Want to join the conversation?

- At5:40, why is it that limits only exist when both sides approach the same value?(18 votes)
- For a good intuitive sense on why this is the case, use your favorite graphing utility and graph the function f(x) = 1 / x. The shape of this particular graph is called a hyperbola.

You might notice that the graph drops rapidly upon reaching as y value of 0 from the negative (or left hand side). You'll also notice that the graph rises rapidly as the y value approaches 0 from the left had side.

Since the graph contains a discontinuity (and a pretty major one at that), the limit of the function as x approaches 0 does not exist, because the 0+ and 0- limits are not equal.(30 votes)

- I don't understand why the x-3 in the numerator becomes negative when x<3. Anyone care to explain?(11 votes)
- You know absolute value of a real number x is non-negative value of x. x can be either negative, zero or positive. So when we take absolute value of x ,(i.e. |x|)

1. when x is zero (x=0), then |x| = |0| = 0

2. when x is positive (x>0), then |x| = positive value

3. when x is negative(x<0), then |x| = -1*x = positive value , so you are getting absolute value of a negative number x and in order to get non-negative same magnitude of x, you multiply negative value of x with -1 and you get positive value.

In the example of this video, you are taking absolute value of x-3, so if

1. x=3, you are getting |3-3|=0

2. x > 3, (x-3) is positive, so |x-3| positive

3. x < 3, (x-3) is negative, so |x-3| = -1*(x-3), which is positive(36 votes)

- isn't there a way to solve these limits without graphs and manual substitution method?(3 votes)
- Absolutely, that's what Calculus is all about :D(26 votes)

- Are "does not exist" and "it's undefined" the same thing?(10 votes)
- They are not. A limit of a function does not exist means it's limits from left and right aren't congruent. An example would be the floor function [x]. When you approach the same number (any point) from the left, you get a value x for it, but when you approach it from the right you get x+1. When it's undefined the limit can be calculated when you approach it from one side, but not the other. 1/x is an undefined function.(6 votes)

- Like a few people I also had troube understanding why the negative is added to the terms describing x<3

If it helps anyone the way I came to terms with it is that the negative is added to enforce the absolute value term of the original function. Any value below 3 will turn out a negative number, which is not an absolute value, so to counter it a -1 is applied/multiplied to the value of the numerator. And of course, because the denominator is still going to turn out a negative number because x<3, we still get a negative number as our final value. Without the absolute value it would turn out a positive 1 instead of a negative.

tbh, I don't understand why we have to go to the trouble of defining that in this case, as the terms are simple enough to understand without it/why are we unable to or not allowed to use the absolute value symbol when defining f(x) when x<3, but I'm assuming it's to form the habit of being used to doing it when we have more complicated problems to address in the future.(6 votes)- There's an even simpler way to put this. The absolute value function basically asks this question: "What do I do to the given number to make it a positive one?"

So, if I have, say, 5, the absolute value won't do anything to it and hence, it remains 5. But, if I had -5, the absolute value would multiply -1 and make it 5.

Also, learning to define the modulus function as a piecewise is really important, especially when you start integration, where a lot of integrals with modulus pop up. There, it is important to know on which interval the function is positive, and on which it's negative.(7 votes)

- For those wondering why the negative absolute value computation gives us a negative number (as I did), remember that there's a denominator that becomes negative! We're so focused on the numerator being an absolute value, we forget to look at the denominator. The numerator is indeed positive, but not the denominator.

In other words :

x>3 --> if x=4, f(4) = 1/1 = 1 (first case)

x<3 --> if x=2, F(2) = 1/-1 = -1 (second case)(8 votes) - I know you can take limit from both positive or negative direction of real numbers but is it possible to approach a limit from the complex numbers plane ?(4 votes)
- For a real function, no, because the function is undefined in the imaginary domain. However, there are such things as complex limits for complex functions.(4 votes)

- why cant the limit be |1|?(0 votes)
- how about the absolute limit of f(x) as x approaches 3? that would be 1, no? (don't actually know if this'd be useful anywhere, but still)(1 vote)

- Why do we say that the limit doesn't exist when the limits from both sides give us different answers? why can't there be two limits or just one limit?(3 votes)
- The limit of a function gives the value of the function as it gets infinitely closer to an x value. If the function approaches 4 from the left side of, say, x=-1, and 9 from the right side, the function doesn't approach any one number. The limit from the left and right exist, but the limit of a function can't be 2 y values.(4 votes)

- how do we define a 'limit' .(2 votes)
- The y value a function approaches when x approaches a certain value. If the y has different values as the x value is approached, then the limit doesn't exist. For example, in the video, as you looked at the graph from the left, you get a value of -1, and from the right, you get 1. So, as -1 is not equal to 1, there is no limit. If they approach the same y value from both sides, the limit does exist, and is is equivalent to the y value approached.(3 votes)

## Video transcript

Let's say that f(x) is equal to the absolute value of x minus three over x minus three and what I'm curious about is the limit of f(x) as x approaches three and just from an inspection you can see that the function is not defined when x is equal to three - you get zero over zero: it's not defined So to answer this question let's try to re-write the same exact function definition slightly differently So let's say f(x) is going to be equal to - and I'm going to think of two cases: I'm going to think of the case when x is greater than three and when x is less than three So when x is - I'll do this in two different colors actually When x - I'll do it in green - that's not green When x is greater than three... When x is greater than three, what does this function simplify to? Well, whatever I get up here, I'm just taking the.. I'm going to get a positive value up here and then I'm... Well, if I take the absolute value it's going to be the exact same thing, so let me... For x is greater than three, this is going to be the exact same thing as x minus three over x minus three because if x is greater than three, the numerator's going to be positive, you take the absolute value of that, you're not going to change its value so you get this right over here or, if we were to re-write it... ...if we were to re-write it, this is equal to, for x is greater than three, you're going to have f(x) is equal to one for x is greater than three Similarly, let's think about what happens when x is less than three When x is less than three, well, x minus three is going to be a negative number When you take the absolute value of that, you're essentially negating it so it's going to be the negative of x minus three over x minus three or if you were to simplify these two things, for any value as long as x doesn't equal three this part right over her simplifies to one, so you are left with a negative one negative one for x is less than three I encourage you, if you don't believe what I just said, try it out with some numbers Try out some numbers: 3.1, 3.001, 3.5, 4, 7 Any number greater than three, you're going to get one You're going to get the same thing divided by the same thing and try values for x less than three: you're going to get negative one no matter what you try So let's visualise this function now So, now you draw some axes... That's my x- axis and then this is my... This is my f(x) axis - y is equal to f(x) and what we care about is x is equal to three so x is equal to one, two, three, four, five and we could keep going... and let's say this is positive one, two, so that's y is equal to one this is y is equal to negative one and negative two and we can keep going... So this way that we have re-written the function is the exact same function as this we've just written [it] in a different way and so what we're saying is... is we're... Our function is undefined at three but if our x is greater than three, our function is equal to one so if our x is greater than three, our function is equal to one so it looks like... It looks like that, and it's undefined at three and if x is less than three our function is equal to negative one so it looks like - I'll be doing that same color It looks like this... It looks like... Looks like this... Once again, it's undefined at three So it looks like that So now let's try to answer our question: What is the limit as x approaches three? Well, let's think about the limit as x approaches three from the negative direction, from values less than three So let's think about first the limit... ...the limit, as x approaches three... ...as x approaches three, the limit of f(x)... ...as x approaches three from the negative direction and all this notation here - I wrote this negative as a superscript right after the three - says Let's think about the limit as we're approaching... ...let me make this clear... Let's think about the limit as we're approaching from the left So in this case, if we get closer... If we get... If we start with values lower than three as we get closer and closer and closer... So, say we start at zero, f(x) is equal to negative one We go to one, f(x) is equal to negative one We go to two, f(x) is equal to negative one If you go to 2.999999, f(x) is equal to negative one So it looks like it is approaching negative one if you approach.. ...if you approach from the left-hand side Now let's think about the limit... ...the limit of f(x)... ...the limit of f(x) as x approaches three from the positive direction, from values greater than three So here we see, when x is equal to five, f(x) is equal to one When x is equal to four, f(x) is equal to one When x is equal to 3.0000001, f(x) is equal to one So it seems to be approaching... It seems to be approaching positive one So now we have something strange We seem to be approaching a different value when we approach from the left than when we approach from the right and if we are approaching two different values then the limit does not exist So this limit right over here does not exist or another way of saying it: The limit... ...the limit of... (Let me write this in a new color - I have a little idea here) ...the limit of a function f(x) as x approaches some value c is equal to L if and only if... ...if and only if the limit of f(x) as x approaches c from the negative direction is equal to the limit of f(x) as x approaches c from the positive direction which is equal to L This did not happen here - the limit when we approached the left was negative one, the limit when we approached from the right was positive one, So we did not get the same limits when we approached from either side So the limit does not exist in this case