amonoconnor
  • amonoconnor
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amonoconnor
  • amonoconnor
\[\lim_{x \rightarrow 2}\frac{ \sqrt{6-x}-2 }{\sqrt{3-x}-1 }\]
anonymous
  • anonymous
You can multiply the bottom by the conjugate or since the value is 0/0, you can use L'Hopital's rule
amonoconnor
  • 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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amonoconnor
  • amonoconnor
\[\frac{\sqrt{18-9x+x^2}+\sqrt{6-x}-2\sqrt{3-x}-2}{(3-x) - 1}\]
amonoconnor
  • amonoconnor
Is that what it should work out to? :/
jim_thompson5910
  • 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
  • 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
  • amonoconnor
I'll show you what I did the first time, and got points off for:
amonoconnor
  • 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
  • amonoconnor
That algebra is in fact wrong, correct, at least the 1st term of the numerator in the product?
jim_thompson5910
  • 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
  • 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
  • amonoconnor
Damn, alright...
amonoconnor
  • amonoconnor
I'm going to take a minute to work through it if that's okay?
jim_thompson5910
  • 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
  • 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
  • amonoconnor
Is this a rule for any situations with a radical in both the numerator and denominator?
jim_thompson5910
  • jim_thompson5910
not that I can remember, but this trick of using 2 conjugates seems to work
amonoconnor
  • 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
  • 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
  • jim_thompson5910
\[\Large (a+b)*(c+d) \ne a*c + b*d\]
jim_thompson5910
  • jim_thompson5910
you have to use the FOIL rule
amonoconnor
  • 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
  • jim_thompson5910
ok go ahead
amonoconnor
  • 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
  • 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
  • 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
  • amonoconnor
Can I cancel out that string of common terms in the top and bottom?
amonoconnor
  • amonoconnor
Sorry, in mine?
jim_thompson5910
  • 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
  • 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
  • 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
  • amonoconnor
That just makes the limit, or answer, go to zero then... that can't be right.
jim_thompson5910
  • 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
  • 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
  • amonoconnor
Yep, that's what I did:)
jim_thompson5910
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • amonoconnor
I don't get what to do then!
jim_thompson5910
  • 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
  • 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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