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.

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Evaluating Limits Algebraically, Part 1 - Concept

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.

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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.

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