##### Like what you saw?

##### Create FREE Account and:

- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- Attend and watch FREE live webinar on useful topics

# One-Sided Limits - Problem 2

FREE###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

You can find some limits by plugging in values of x into a function and seeing what number the function seems to be getting closer and closer to as the values of x approach a certain point, call it p. Start with finding the left-hand limit. To find what f(x) is as it approaches point p from the left, plug in values of x that are less than p. As the values of x get closer and closer to p, look to see what number f(x) gets closer to. To find the right-hand limit, do the same thing, except instead of using values of x that are less than p, use values of x that are greater than p. Then, you will be able to see what f(x) is as it approaches p. If the left-hand and right-hand limits are not the same, then the limit as x approaches a of f(x) does not exist.

Let’s do another problem. I want to find the limits of this function g(x) equals the absolute value of x² minus 1 over x plus 1. First the limit as x approaches -1 from the left. Let’s look at the TI 84.

Here we are on the TI 84; let me first enter my function; y equals. Now to get absolute value, you can get it a couple of ways. One is to go into the math menu, under numbers. If you can’t remember that, you can always find it in the catalogue, second zero. It’s the very first entry in the catalogue and that’s how I usually like to get it.

Absolute value of x² minus 1 divide by x plus 1. And now we want to go into table set. Let’s make sure that, so the independent variable set to ask, dependent variable set to auto. So I can enter x values myself and the calculator will tell me what the y values are. Let’s go to the table.

I want x to approach -1 from the left. So a number to the left of -1 is -2 and I enter. I get -3. Let’s get a little closer to -1, -1.1, I get -2.1. How about -1.01? I get -2.01. I’ll do it one more time. -1.001, I get -2.001. It looks like these values are headed toward -2. Let me write these on the board. I’ve got my values written on the board here. As I said before, we have x going towards -1 from the left. So -2, -1.1, -1.01, and so on, approaching -1 from the left.

What are the y values doing? It looks like they’re headed towards -2. So I would say as the limit approaches -1 from the left of g(x) is -2. Now let’s take a look as the limit as x approaches -1 from the right of g(x). We use the TI 84 again. We’re back on the TI 84. Now this time I want to approach -1 from the right. So let me start at a value to the right like 0.

When I plug in 0 I get 1. Now let me try -0.9, I get 1.9. Let me try -0.99, I get 1.99. This is interesting, so -0.999. This is interesting because before the y values were approaching -2, this time looks like they are approaching +2 as x gets closer and closer to -1. So let’s record this on the board.

I’ve recorded the values on the board. I have x approaching -1, this time from the right. Now remember zero is to the right of -1 so we’re coming in from the right side on the number line. 0, -.9, -.99 and so on. As these numbers head towards -1, what are the y values doing? 1, 1.9, 1.99, it looks like they’re headed towards 2. I would say the limit as x approaches -1 from the right of g(x) is 2.

Now let’s take a look, this value was 2, but this other value, the limit as x approaches -1 from the left was -2. So we can conclude, for part c that the limit approaches -1 of g(x) does not exist. The only way for that limit to exist is for both one sided limits to exist and to be equal, and they are not equal. So we say the two sided limit as x approaches -1, of g(x) does not exist.

Please enter your name.

Are you sure you want to delete this comment?

###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

###### Get Peer Support on User Forum

Peer helping is a great way to learn. Join your peers to ask & answer questions and share ideas.

##### Concept (1)

##### Sample Problems (3)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete