Name: 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）
Uploaded: 2022-07-05T13:48:25.000Z
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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,

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

x equals 5, the value of the function is equal to c.

out what the limit of f of x as x approaches 5 actually is.

into the expression right up here, in the numerator

And then you get 5 minus 5 in the denominator,

So you get this indeterminate form of 0/0.

do have a tool that allows us, or gives us an option

But we can actually tackle this with a little bit

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,

plus 3 over the square root of x plus 4 plus 3.

We obviously have to multiply the numerator

so that we actually don't change the value of the expression.

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.

that we use often times to get rid of complex numbers,

But if you multiply this out-- and I encourage you to do it--

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.

going to have x minus 5 times the square root of x

And so this has-- I guess you could say simplified to,

algebraically to see if we can then substitute x equals 5

or if we can somehow simplify it to figure out

And when you simplify the numerator up here,

Well, that's x minus 5 over x minus 5 times the square root

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

You can divide the numerator and the denominator by x minus 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.

and this is a simpler expression-- with 1

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,

We can use this expression right over here.

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.

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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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