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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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11-2 尋找極限並使函數連續的花式代數(Fancy algebra to find a limit and make a function continuous | Differential Calculus | Khan Academy)

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    yukang920108 發佈於 2022 年 07 月 05 日
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