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## AP®︎/College Calculus AB

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

Lesson 6: Determining limits using algebraic properties of limits: direct substitution

# 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?

• At , why is it that limits only exist when both sides approach the same value? •  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.
• I don't understand why the x-3 in the numerator becomes negative when x<3. Anyone care to explain? •  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
• isn't there a way to solve these limits without graphs and manual substitution method? • Are "does not exist" and "it's undefined" the same thing? • 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.
• 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) • 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. • 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.
• 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 ? • why cant the limit be |1|?   