Question

Limit as x approaches 9. (Sqrt(x)-3)/(x-9)

Answer 1

\[\lim_{x \rightarrow 9} \frac{\sqrt{x-3}}{x-9}\] is this it?

Answer 2

Or is it \[\lim_{x \rightarrow 9} \frac{\sqrt{x}-3}{x-9}\]

Answer 3

its the second one

Answer 4

sorry i was on my phone and it didnt let me reply and i couldnt remember my password

Answer 5

What is another way you can write out `x-9`?

Answer 6

Remember, this is a difference of squares.

Answer 7

\[x = \sqrt{x} \cdot \sqrt{x}\]\[9 = 3 \cdot 3\]

Answer 8

And difference of squares, \((a-b)^2 = (a-b)(a+b)\)

Answer 9

so it's \[(\sqrt{x}-3)^2\]

Answer 10

?

Answer 11

Yep :)

Answer 12

then id cancel out one and id be left with 1/sqrt(9)+3 making it 6 so it'd be 1/6

Answer 13

awesome thank you :)

Answer 14

So you would have : \[\lim_{x \rightarrow 9} \frac{\sqrt{x}-3}{(\sqrt{x}-3)^2}\]

Answer 15

This can be simplified how?

Answer 16

by separating the bottom?

Answer 17

Mmhmm, and what does that factor out to according to the difference of squares?

Answer 18

\[(\sqrt{x}-3)(\sqrt{x}-3) \]

Answer 19

?

Answer 20

wait no a plus in one of them?

Answer 21

Yep, onehas a plus.

Answer 22

So therefore you would have \[\lim_{x \rightarrow 9} \frac{\color{red}{\sqrt{x}-3}}{(\color{red}{\sqrt{x}-3})(\sqrt{x}+3)}\]

Answer 23

Do you see how the ones highlighted in red can be reduced to 1?

Answer 24

\[\frac{ 1 }{ \sqrt{9}-3 }\]

Answer 25

yes, then it'd be that

Answer 26

\[\frac{ 1 }{ 6 }\]

Answer 27

NOt quite, theres an error in your sign

Answer 28

whats the error?

Answer 29

OH the plus sign

Answer 30

\(\color{blue}{\text{Originally Posted by}}\) @Calcisnotforme2_
\[\frac{ 1 }{ \sqrt{9}\color{red}{-}3 }\]
\(\color{blue}{\text{End of Quote}}\)

Answer 31

just a quick check

\[(a - b)^2 \neq(a -b)(a+b)\]

as stated earlier

Answer 32

it should be a plus sign*

Answer 33

Oh, true campbell, thanks for identifying that. Rather, it is \(a^2-b^2 = (a-b)(a+b)\)

Answer 34

And yes @Calcisnotforme2_ it's a plus sign

Answer 35

im just confused now after the note from campbell

Answer 36

Correction from earlier: \((a-b)^2 = a^2-2ab+b^2\), sorry if that confused you.

Answer 37

But x-9 isn't squared. that is what im confused about. to get it to factor the way we factored it.

Answer 38

Let me clarify the two identities. and then i'll clarify which one i used to simplify the denominator.

Answer 39

1. \((a-b)^2 = \sf \text{difference of squares} = a^2 -2ab+b^2\)

2. \(\sf a^2 - b^2 = (a-b)(a+b)\)

Answer 40

If you expand the second one, you will end up with \(\sf a^2 -ab +ab - b^2 \iff a^2 -b^2\)

Answer 41

Therefore, we use the second identity to solve for \(\sf x-9\) since it can be rewritten as \(\sf (\sqrt{x})^2 -(3)^2\).

Answer 42

Does that clarify things a bit more?

Answer 43

Oh okay i get it now, thank you :)

Answer 44

Yay :) You're welcome :D