Calculate the Limit Algebraically

hopelynnwelch

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Limit as x goes to 9 of (sqrt(x) - 3)/(x-9)

To figure this out algebraically I would sub 9 in for x but I can't because that would make the function undefined. So I guess what I think I should do is multiply the entire function by the conjugate of the numerator.

That leaves me with (x-9)/(x(sqrt(x))+3x-9(sqrt(x))-27)

But if I sub 9 into this, my function is still undefined.

I could make a table and sub in numbers close to 9 from a negative and positive direction. For example 8.9, 8.99, 8.999 and 9.1, 9.01, 9.001.

But would that be calculating algebraically? Is there some way of manipulating this function I am just not seeing?

It has been 5 years since I took Functions for Calc and I am now taking calc I and II (I had babies) So my algebra is pretty rusty...
 
Limit as x goes to 9 of (sqrt(x) - 3)/(x-9)

To figure this out algebraically I would sub 9 in for x but I can't because that would make the function undefined 0/0 is indetermined, not undefined. So I guess what I think I should do is multiply the entire function by the conjugate of the numerator.

That leaves me with (x-9)/(x(sqrt(x))+3x-9(sqrt(x))-27) do not multiply out!. The x-9 in the denominator will cancel out the x-9 in the numerator leaving 1/(sqrt(x) +3). This goes to 1/6 as x goes to 9

But if I sub 9 into this, my function is still undefined. If a limit is undefined then you are done, the limit does not exist. But again 0/0 is NOT undefined

I could make a table and sub in numbers close to 9 from a negative and positive direction. For example 8.9, 8.99, 8.999 and 9.1, 9.01, 9.001.

But would that be calculating algebraically? Is there some way of manipulating this function I am just not seeing?

It has been 5 years since I took Functions for Calc and I am now taking calc I and II (I had babies) So my algebra is pretty rusty...
Your algebra is fine! You want to get rid of the 0/0, so do not multiply out the conjugate in the denominator with (x-9)
 
Limit as x goes to 9 of (sqrt(x) - 3)/(x-9)

To figure this out algebraically I would sub 9 in for x but I can't because that would make the function undefined. So I guess what I think I should do is multiply the entire function by the conjugate of the numerator.

That leaves me with (x-9)/(x(sqrt(x))+3x-9(sqrt(x))-27)

But if I sub 9 into this, my function is still undefined.

I could make a table and sub in numbers close to 9 from a negative and positive direction. For example 8.9, 8.99, 8.999 and 9.1, 9.01, 9.001.

But would that be calculating algebraically? Is there some way of manipulating this function I am just not seeing?

It has been 5 years since I took Functions for Calc and I am now taking calc I and II (I had babies) So my algebra is pretty rusty...

Note that
a2 - b2 = (a-b ) (a+b)
x = (x\displaystyle \sqrt{x})2
9 = 32
 
a2 - b2 = (a-b ) (a+b)
x = (x\displaystyle \sqrt{x})2
9 = 32[/QUOTE]

I wish I could say that helped me but I might not be seeing where you are going with it. I do understand what Jomo is saying and how if I don't multiply the denominator I can cancel... So I'm going to run with that.
 
I wish I could say that helped me but I might not be seeing where you are going with it. I do understand what Jomo is saying and how if I don't multiply the denominator I can cancel... So I'm going to run with that.

Let
a = x\displaystyle \sqrt{x}
b = 3
Then go one of several ways one of which is
x3x9=x3x9x+3x+3\displaystyle \frac{\sqrt{x} -3}{x - 9} = \frac{\sqrt{x} -3}{x - 9} \frac{\sqrt{x} + 3}{\sqrt{x} + 3}
= (x3)(x+3)(x9)(x+3)\displaystyle \frac{(\sqrt{x} -3) (\sqrt{x} + 3)}{(x - 9) (\sqrt{x} + 3)}
= x9(x9)(x+3)\displaystyle \frac{x - 9}{(x - 9) (\sqrt{x} + 3)}
= 1x+3\displaystyle \frac{1}{\sqrt{x} + 3}
or another
x3x9=x3(x3)(x+3\displaystyle \frac{\sqrt{x} -3}{x - 9} = \frac{\sqrt{x} -3}{(\sqrt{x} - 3) (\sqrt{x} + 3}
= 1x+3\displaystyle \frac{1}{\sqrt{x} + 3}
 
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