##### 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

# Evaluating Limits Algebraically, Part 1 - Concept

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.

When we evaluate a limit, we are trying to determine the value that the function is approaching at a certain point. When **evaluating limits**, we want to first check to see if the function is continuous. If we determine that the limit is continuous at the point where we are evaluating it, we can simply plug in the value and solve the function.

One of the first things that you will learn to do in a Calculus class is how to evaluate a limit, and I want to show you how to evaluate limits Algebraically. Now I've divided evaluating limits Algebraically into two categories, sometimes you can use continuity and sometimes you can't. So I'm going to go over the cases where you can use continuity and uh and then later I'll go over the ones where you can't.

Let's start with the definition of continuity, let's remind ourselves, a function of is continuous at the point x=a then limit as x approaches a of f of x equals f of a. And what this means is when you're evaluating limits you can basically plug in the value as expert you say you can plug in the a value into f of x and that's the value of your limit. You can only do this for continuous functions but luckily a lot of the functions we deal with are continuous. Here's an example this looks like a terrible limit limit is x approaches -2 of x squared plus 8x-20 over x-2. Now this is a rational function so it's going to be continuous everywhere it's defined. And it's only undefined for x=+2 right, so this is continuous for x not equal to 2. Well we're letting x approach -2 we're pretty clear away from positive 2. So I can evaluate this limit by plugging in, so this limit is going to equal -2 squared +8x-2-20 over -2-2. So I will just have to do this Arithmetic right, I have 4-16 -20 and I have -2-2 -4. Now what do we have on top here? 4-36 -32 over -4 this is 8. And so the value of the limit is 8 right?

Now here I use continuity, you use continuity whenever you have a continuous function you just plug in the value of the number a whatever x is approaching into the function and evaluate it. Now if that doesn't seem to work like if you get division by 0 when you do that then the function is not continuous at that point and you have to use some other method.

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.

##### 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