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# Removing discontinuities (rationalization)

AP.CALC:
LIM‑2 (EU)
,
LIM‑2.C (LO)
,
LIM‑2.C.1 (EK)
,
LIM‑2.C.2 (EK)

## 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 F if X does not equal 5 and it's equal to C if x equals 5 and 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 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 over 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 limits when we get this indeterminate form it's called a la Patel'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 anytime you see a radical plus or minus something else to get rid of the radical what you can do is multiply by the radical doing the 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 and this is a technique that we learn in algebra or sometimes in precalculus 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 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 is the difference of squares something minus something times something plus something so the first term is going to be the first two something squared so square root of x plus four squared is X plus four 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 three squared so minus nine and then in the denominator you're of course going to have X minus five X minus five times the square root of x plus four plus three and so this has I guess you could say simplified to although it's not arguably any simpler but at least we've gotten our radical we're early just playing around with it algebraically to see if we can then substitute x equals five 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 four minus nine well that's X minus 5 over X minus five times the square root of x plus four plus three and now it pops out at you both the numerator and the denominator are now divisible by X minus five 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 five 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 for X does not equal 5 which is fine because in this first the first part of this function definition this is for the case where X does not equal 5 so we could actually replace this and this is 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 5 where you 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 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
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