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Removing discontinuities (rationalization)

Sal finds the value the function f(x)=(√(x+4)-3)/(x-5) should have at x=5 so it's continuous at that point. Created by Sal Khan.
Video transcript
Let f be the function given by f of x is equal to the square root of x plus 4 minus 3 over x minus 5. If x does not equal 5, and it's equal to c if x equals 5. Then say, if f is continuous at x equals 5, what is the value of c? So if we know that f is continuous at x equals 5, that means that the limit as x approaches 5 of f of x is equal to f of 5. This is the definition of continuity. And they tell us that f of 5, when x equals 5, the value of the function is equal to c. So this must be equal to c. So what we really need to do is figure out what the limit of f of x as x approaches 5 actually is. Now, if we just try to substitute 5 into the expression right up here, in the numerator you have 5 plus 4 is 9. The square root of that is positive 3, the principal root is positive 3. 3 minus 3 is 0. So you get a 0 in the numerator. And then you get 5 minus 5 in the denominator, so you get 0 in the denominator. So you get this indeterminate form of 0/0. And in the future, we will see that we do have a tool that allows us, or gives us an option to attempt to find the limits when we get this indeterminate form. It's called L'Hopital's rule. But we can actually tackle this with a little bit of fancy algebra. And to do that, I'm going to try to get this radical out of the numerator. So let's rewrite it. So we have the square root of x plus 4 minus 3 over x minus 5. And any time you see a radical plus or minus something else, to get rid of the radical, what you can do is multiply by the radical-- or, if you have a radical minus 3, you multiply by the radical plus 3. So in this situation, you just multiply the numerator by square root of x plus 4 plus 3 over the square root of x plus 4 plus 3. We obviously have to multiply the numerator and the denominator by the same thing so that we actually don't change the value of the expression. If this right over here had a plus 3, then we would do a minus 3 here. This is a technique that we learn in algebra, or sometimes in pre-calculus class, to rationalize usually denominators, but to rationalize numerators or denominators. It's also a very similar technique that we use often times to get rid of complex numbers, usually in denominators. But if you multiply this out-- and I encourage you to do it-- you notice this has the pattern that you learned in algebra class. It's a difference of squares. Something minus something times something plus something. So the first term is going to be the first something squared. So square root of x plus 4 squared is x plus 4. And the second term is going to be the second something, or you're going to subtract the second something squared. So you're going to have minus 3 squared, so minus 9. And in the denominator, you're of course going to have x minus 5 times the square root of x plus 4 plus 3. And so this has-- I guess you could say simplified to, although it's not arguably any simpler. But at least we have gotten our radical. We're really just playing around with it algebraically to see if we can then substitute x equals 5 or if we can somehow simplify it to figure out what the limit is. And when you simplify the numerator up here, you get x plus 4 minus 9. Well, that's x minus 5 over x minus 5 times the square root of x plus 4 plus 3. And now it pops out at you, both the numerator and the denominator are now divisible by x minus 5. So you can have a completely identical expression if you say that this is the same thing. You can divide the numerator and the denominator by x minus 5 if you assume x does not equal 5. So this is going to be the same thing as 1 over square root of x plus 4 plus 3 for x does not equal 5. Which is fine, because in the first part of this function definition, this is the case for x does not equal 5. So we could actually replace this-- and this is a simpler expression-- with 1 over square root of x plus 4 plus 3. And so now when we take the limit as x approaches 5, we're going to get closer and closer to five. We're going to get x values closer and closer to 5, but not quite at 5. We can use this expression right over here. So the limit of f of x as x approaches 5 is going to be the same thing as the limit of 1 over the square root of x plus 4 plus 3 as x approaches 5. And now we can substitute a 5 in here. It's going to be 1 over 5 plus 4 is 9, principal root of that is 3. 3 plus 3 is 6. So if c is equal to 1/6, then the limit of our function as x approaches 5 is going to be equal to f of 5. And we are continuous at x equals 5. So it's 1/6.