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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.