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• 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.

Let f be the function given by f of x

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# Fancy algebra to find a limit and make a function continuous | Differential Calculus | Khan Academy

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yukang920108 posted on 2022/07/05
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