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!

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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!

Calculus1
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\[\lim_{x \rightarrow 2}\frac{ \sqrt{6-x}-2 }{\sqrt{3-x}-1 }\]
You can multiply the bottom by the conjugate or since the value is 0/0, you can use L'Hopital's rule
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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\[\frac{\sqrt{18-9x+x^2}+\sqrt{6-x}-2\sqrt{3-x}-2}{(3-x) - 1}\]
Is that what it should work out to? :/
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
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! :(
I'll show you what I did the first time, and got points off for:
\[\lim_{x \rightarrow 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}*(\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}) = \lim_{x \rightarrow 2}\frac{\sqrt{9+x^2}+\sqrt{6-x}-2\sqrt{3-x}-2}{\sqrt{9+x^2}-1}\]
That algebra is in fact wrong, correct, at least the 1st term of the numerator in the product?
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
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 \[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]
Damn, alright...
I'm going to take a minute to work through it if that's okay?
notice this pairing here \[\Large \frac{\color{red}{\sqrt{6-x}-2}}{\sqrt{3-x}-1}\times\frac{\color{red}{\sqrt{6-x}+2}}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]
and this pairing as well \[\Large \frac{\sqrt{6-x}-2}{\color{red}{\sqrt{3-x}-1}}\times\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\color{red}{\sqrt{3-x}+1}}\]
Is this a rule for any situations with a radical in both the numerator and denominator?
not that I can remember, but this trick of using 2 conjugates seems to work
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)?
\[\Large \sqrt{6-x}*\sqrt{3-x} = \sqrt{(6-x)*(3-x)}\] \[\Large \sqrt{6-x}*\sqrt{3-x} = \sqrt{18 - 9x + x^2}\]
\[\Large (a+b)*(c+d) \ne a*c + b*d\]
you have to use the FOIL rule
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 ?
ok go ahead
\[\lim_{x \rightarrow 2}\frac{ \sqrt{6-x}-2 }{\sqrt{3-x}-1}*\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}*\frac{ \sqrt{3-x}+1 }{\sqrt{3-x}+1} = \lim_{x \rightarrow 2}\frac{(6-x)-4+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}{(3-x)-1+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}\]
\[\lim_{x \rightarrow 2}\frac{(6-x)-4+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}{(3-x)-1+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}\]
why do all that work? Seems way too tedious than it needs to be. Here's how I did it \[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\] \[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\] \[\Large \frac{6-x-4}{\sqrt{3-x}-1}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\] \[\Large \frac{6-x-4}{3-x-1}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\] \[\Large \frac{2-x}{2-x}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\] \[\Large \frac{1}{1}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\]
Can I cancel out that string of common terms in the top and bottom?
Sorry, in mine?
`Can I cancel out that string of common terms in the top and bottom?` you mean like canceling the x terms in something like \[\Large \frac{x+3}{x+2}\]
I'm not sure what you mean. Have a look at my steps. Let me know if they make sense or not
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?
That just makes the limit, or answer, go to zero then... that can't be right.
no you can't do this \[\lim_{x \rightarrow 2}\frac{(6-x)-4+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}{(3-x)-1+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}\] \[\lim_{x \rightarrow 2}\frac{(6-x)-4+\cancel{\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}}{(3-x)-1+\cancel{\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}}\]
you agree with this step right? \[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]
Yep, that's what I did:)
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?
Correct:) I think I got the multiplying across right, but I just don't know if you can break it down more :/
ok so after doing that, we will get to this step \[\Large \frac{2-x}{\sqrt{3-x}-1}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]
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 \[\Large \frac{2-x}{2-x}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\]
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?
I don't think there is a way to simplify that result you got, which is why I avoided expanding things out
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...
I don't get what to do then!
do you see how I got to \[\Large \frac{2-x}{2-x}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\]
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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