Brightstorm is like having a personal tutor for every subject
See what all the buzz is aboutCheck it out
Evaluating Limits Algebraically, Part 2 - Problem 2 3,721 views
Previously, we learned that you can evaluate the limit of a function when a function is continuous (and therefore defined) at the point you are evaluating. However, when a function is not continuous at that point, something different must be done. For rational functions (functions with polynomials as the numerator and denominator), the function is not defined at any point where the denominator is 0. For example, if the function is f(x)=(x^2+2x-15)/(x-3), the function would be undefined when x=3 because 3-3=0. When you have a rational function that is undefined at the point at which you want to find the limit, check to see if it is possible to factor the numerator and/or denominator. By factoring, it may be possible to cancel out terms. This may help by canceling out the factor with the undefined point. Going back to the example, factor the numerator and denominator. The numerator becomes (x+5)(x-3), and the denominator is already factored as x-3. The x-3 terms cancel so the function becomes x+5, which is defined at x=3. As a result of factoring and canceling, you can evaluate the limit by plugging in the value of x at that point, because f(x) is now defined there.
Let’s do another example of a limit. Evaluating a limit algebraically, when continuity doesn’t work. So here I have one. Limit whose x approach -3 of x over x plus 4 plus 3, over x plus 3. Now you can see in the situation, as x approaches -3 my denominator is going to 0. So this is why I can’t use continuity.
So I have to somehow use algebra to eliminate that problem of division by 0. So let’s do that. Limit as x approaches -3. As I manipulate the expression inside the limit, remember you have to keep writing the limit expression down, until you actually evaluate it. So I’m going to write limit as x approaches -3. And then on the top I’m going to multiply the top and bottom by x plus 4.
This algebraic trick is going to get rid of the little denominator here. Right now I have a complex fraction and I want to make it a simple fraction. It will be easier to evaluate. X plus 4 times x over x plus 4, is x and 3 times x plus 4 is 3x plus 12.
In the denominator I have x plus 3 times x plus 4 and it's actually better that I leave that factor. Because remember, my goal here is to cancel out the x plus 3 if possible. That’s the problem factor that’s the one that goes to 0 as x goes to -3. So don’t multiply this through. I know it's tempting.
Although you have a numerator, this numerator, this is 4x plus 12, which is the same as 4 times the quantity x plus what? 3. Perfect. So the x plus 3 on the top will cancel with the x plus 3 in the bottom. And that will get rid of the problem. So let me cancel those and what I have is the limit as x approaches -3, a 4 over x plus 4.
So as x approaches -3, the top is going to 4 and the bottom is going to -3 plus 4, 1. So my limit is 4. Remember, whenever you have to evaluate a limit algebraically and continuity doesn’t work, the problem is going to be division by a 0. And if your teacher expects you to evaluate that limit there has to be some algebraic way of eliminating that problem.
So look for ways to factor the numerator and denominator and cancel factors.