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# Evaluating Limits Algebraically, Part 2 - Problem 1

###### 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.

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 take a look at an example of a limit where we can’t use continuity. Here’s a problem that says evaluate the limit as x approaches -4 of x² minus 16, over x² plus 4x. The reason I can’t use continuity here is that this rational function is not defined at -4. -4 will give me 16 minus 16 or 0. So I can’t just plug in -4. What I have to do is, I have to look at this rational function and see how it factors. Look at the numerator and the denominator.

Now the numerator factor is in to x minus 4 x plus 4, because it’s a difference of squares. And the denominator as a common factor of x. So I can pull that out and that leaves x plus 4. And now I see that the factor that causes the problem, is the x plus 4. That’s the one that goes to 0, as x approaches -4 and it cancels.

So when you are evaluating limits, you can cancel the problem factors. This gives me the limit as x approaches -4 of x minus 4 over x. And now with this remaining piece of the function, as x approaches -4, I don’t have any problems with divisions by 0.

I can just plug in -4. I can use continuity to evaluate. And I get -4 minus 4, -8, over -4. Notice I’m evaluating the limit, and so I don’t have to write limit any more. At this point of evaluate the limit, it's -8 over -4 which is just simplified to 2.

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###### 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.

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