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Evaluating Limits Algebraically, Part 2 - Problem 3 3,365 views
When a function is continuous at a point, you can evaluate the limit as x approaches that point by simply evaluating the function at that point. However, with radical functions, the function is not defined everywhere--whatever is inside the radical cannot be negative. Additionally, remember that for rational functions, the denominator cannot be equal to 0. When there is a radical in the denominator, multiply the numerator and denominator by the conjugate of the denominator. So, if the denominator is radical (x+3), then multiply the numerator and denominator by radical (x-3). Look to see if there are factors that can now be cancelled. When you can cancel certain factors, you will be able to evaluate the function in order to find its limit at the desired point, just as we've seen in other limit-finding problems.
Let’s do a tougher problem. I want to evaluate the limit as x approaches 25 of 2x minus 50 over root x minus 5. You can see that this is going to be a problem because, as x approaches 25, root x is approaching 5. So I get 5 minus 5 0 on the denominator. I have to find some way to get rid of this.
When you have a radical, like this, a radical expression root x minus, 5 you have to use a trick which I’ll show you in a second. First, let me just rewrite this as, you look at the numerator. I can pull a 2 out of this numerator and I’m left with x minus 25.
I still have root x minus 5 in the denominator. Here’s the trick. You want to multiply the top and bottom by the conjugate of the denominator. The conjugate of the denominator is root x plus 5. The reason conjugates are so special is, when you multiply an expression like this by its conjugate, you get rid of the radical.
Let’s see what this gives us. Limit as x approaches 25. Now I’m not going to multiply this through. I’m just going to leave it as 2 times x minus 25 times root x plus 5. And then in the denominator, this is going to give me a difference of squares. It's root x quantity squared which is just x, and then minus 5 times 5 -25.
Look at that, now I can see that I cancel this. So it’s good that I didn’t multiply this through. And canceling these guys, I’m left with something that I can plug 25 into. So this is going to be limit as x approaches 25 of 2 times root x plus 5.
So plugging in, I get 2 times root 25 plus 5. That’s 2 times 5, 10 plus 5 so the answer is 15. So remember, even though this was tricky, we eventually used the same trick of eliminating the factor that cause the problem.
The factor that goes to 0 in the denominator. And the way we did was recognize that we had a radical expression in the denominator. We multiplied the top and bottom by its conjugate, that’s the trick for getting rid of one of these differences, that includes a radical.