I'm stuck on this one, and would LOVE some help in understanding the steps in solving it!
(Equation format problem in comment)
Thank you very much!

- amonoconnor

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- amonoconnor

\[\lim_{x \rightarrow 2}\frac{ \sqrt{6-x}-2 }{\sqrt{3-x}-1 }\]

- anonymous

You can multiply the bottom by the conjugate or since the value is 0/0, you can use L'Hopital's rule

- amonoconnor

I multiplied the function by the conjugate of the denominator, but I am not sure if my algebra is accurate for multiplying the numerator across... Can you work through that with me?

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## More answers

- amonoconnor

\[\frac{\sqrt{18-9x+x^2}+\sqrt{6-x}-2\sqrt{3-x}-2}{(3-x) - 1}\]

- amonoconnor

Is that what it should work out to? :/

- jim_thompson5910

You did everything correctly. Unfortunately, if you plugged in x = 2, you'd still get 0/0. You're better off using L'Hospital's rule

- amonoconnor

Is there a way to simplify it further? My professor says that we don't need L'Hospital's Rule (haven't learned it yet, and he wants us to not get ahead of him), though I vaguely remember it from high school. I'm doing quiz corrections, and the first time through I did get a final answer of 2, though think I messed up this algebra I'm trying to get help on.. I can show you what I did the first time, if you want me to. But... If this IS the CORRECT algebra, what you responded to just above in the feed... then how can we actually solve it, without L'Hospital's Rule? Ahh! I'm so confused! :(

- amonoconnor

I'll show you what I did the first time, and got points off for:

- amonoconnor

\[\lim_{x \rightarrow 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}*(\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}) = \lim_{x \rightarrow 2}\frac{\sqrt{9+x^2}+\sqrt{6-x}-2\sqrt{3-x}-2}{\sqrt{9+x^2}-1}\]

- amonoconnor

That algebra is in fact wrong, correct, at least the 1st term of the numerator in the product?

- jim_thompson5910

ok I figured it out. You were on the right path by using the conjugate. But you have to use both the numerator conjugate and the denominator conjugate at the same time

- jim_thompson5910

instead of saying
\[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\]
or
\[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]
you have to do this
\[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]

- amonoconnor

Damn, alright...

- amonoconnor

I'm going to take a minute to work through it if that's okay?

- jim_thompson5910

notice this pairing here
\[\Large \frac{\color{red}{\sqrt{6-x}-2}}{\sqrt{3-x}-1}\times\frac{\color{red}{\sqrt{6-x}+2}}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]

- jim_thompson5910

and this pairing as well
\[\Large \frac{\sqrt{6-x}-2}{\color{red}{\sqrt{3-x}-1}}\times\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\color{red}{\sqrt{3-x}+1}}\]

- amonoconnor

Is this a rule for any situations with a radical in both the numerator and denominator?

- jim_thompson5910

not that I can remember, but this trick of using 2 conjugates seems to work

- amonoconnor

As a rule, do you multiply the stuff under a radical, when multiplying two radicals together, or add it? So like, does the square root of (6-x) * sqrt.(3-x) = sqrt(18-9x+x^2) ? or is it sqrt.(9-x^2)?

- jim_thompson5910

\[\Large \sqrt{6-x}*\sqrt{3-x} = \sqrt{(6-x)*(3-x)}\]
\[\Large \sqrt{6-x}*\sqrt{3-x} = \sqrt{18 - 9x + x^2}\]

- jim_thompson5910

\[\Large (a+b)*(c+d) \ne a*c + b*d\]

- jim_thompson5910

you have to use the FOIL rule

- amonoconnor

I know this might seem tedious, but I just really need to brush up on my algebra... I took Calc as a junior in high school and I couldn't get AP Calc or ITV Calc as a senior, so I had a gap year... the effects are real!
Can you check this for me? @jim_thompson5910 ?

- jim_thompson5910

ok go ahead

- amonoconnor

\[\lim_{x \rightarrow 2}\frac{ \sqrt{6-x}-2 }{\sqrt{3-x}-1}*\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}*\frac{ \sqrt{3-x}+1 }{\sqrt{3-x}+1} = \lim_{x \rightarrow 2}\frac{(6-x)-4+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}{(3-x)-1+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}\]

- amonoconnor

\[\lim_{x \rightarrow 2}\frac{(6-x)-4+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}{(3-x)-1+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}\]

- jim_thompson5910

why do all that work? Seems way too tedious than it needs to be. Here's how I did it
\[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\]
\[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]
\[\Large \frac{6-x-4}{\sqrt{3-x}-1}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]
\[\Large \frac{6-x-4}{3-x-1}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\]
\[\Large \frac{2-x}{2-x}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\]
\[\Large \frac{1}{1}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\]

- amonoconnor

Can I cancel out that string of common terms in the top and bottom?

- amonoconnor

Sorry, in mine?

- jim_thompson5910

`Can I cancel out that string of common terms in the top and bottom?`
you mean like canceling the x terms in something like
\[\Large \frac{x+3}{x+2}\]

- jim_thompson5910

I'm not sure what you mean. Have a look at my steps. Let me know if they make sense or not

- amonoconnor

No, they don't to be honest.. :/ I'm good with solid algebraic rules, once I get refamiliarized with them though:) Is it correct to assume that I CANNOT cancel out everything from the "+" on the top after 4, and the "+" on the bottom after the 1? Should I take the limit of the numerator and denominator separately from here?

- amonoconnor

That just makes the limit, or answer, go to zero then... that can't be right.

- jim_thompson5910

no you can't do this
\[\lim_{x \rightarrow 2}\frac{(6-x)-4+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}{(3-x)-1+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}\]
\[\lim_{x \rightarrow 2}\frac{(6-x)-4+\cancel{\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}}{(3-x)-1+\cancel{\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}}\]

- jim_thompson5910

you agree with this step right?
\[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]

- amonoconnor

Yep, that's what I did:)

- jim_thompson5910

ok if we multiply out the \(\Large \sqrt{6-x}-2\) and \(\Large \sqrt{6-x}+2\), we get \(\Large (6-x)-4 = 2-x\) agreed?

- amonoconnor

Correct:) I think I got the multiplying across right, but I just don't know if you can break it down more :/

- jim_thompson5910

ok so after doing that, we will get to this step
\[\Large \frac{2-x}{\sqrt{3-x}-1}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]

- jim_thompson5910

if we multiply out the \(\Large \sqrt{3-x}-1\) and \(\Large \sqrt{3-x}+1\) we get \(\Large (3-x)-1 = 2-x\)
so we have this next step
\[\Large \frac{2-x}{2-x}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\]

- amonoconnor

I this like a different way to multiply the three fractions? Like a way to not have the mumbo jumbo you said was tedious above? Is there a way to simply the long product?

- jim_thompson5910

I don't think there is a way to simplify that result you got, which is why I avoided expanding things out

- amonoconnor

Gotcha... I never learned any other way, like no tricks or anything, my prof just scribes these long as heck strings of terms, like I have, and then gets a beautiful answer after simplifying, doing this and that, etc...

- amonoconnor

I don't get what to do then!

- jim_thompson5910

do you see how I got to \[\Large \frac{2-x}{2-x}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\]

- amonoconnor

To be honest sir... no, I don't :/ I'll try to find a tutor or something, because you've been extremely patient, and I think it's going to take more than even the OS Gods to help me out here!

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