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amonoconnor

  • one year ago

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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  1. amonoconnor
    • one year ago
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    \[\lim_{x \rightarrow 2}\frac{ \sqrt{6-x}-2 }{\sqrt{3-x}-1 }\]

  2. anonymous
    • one year ago
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    You can multiply the bottom by the conjugate or since the value is 0/0, you can use L'Hopital's rule

  3. amonoconnor
    • one year ago
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    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?

  4. amonoconnor
    • one year ago
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    \[\frac{\sqrt{18-9x+x^2}+\sqrt{6-x}-2\sqrt{3-x}-2}{(3-x) - 1}\]

  5. amonoconnor
    • one year ago
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    Is that what it should work out to? :/

  6. jim_thompson5910
    • one year ago
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    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

  7. amonoconnor
    • one year ago
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    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! :(

  8. amonoconnor
    • one year ago
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    I'll show you what I did the first time, and got points off for:

  9. amonoconnor
    • one year ago
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    \[\lim_{x \rightarrow 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}*(\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}) = \lim_{x \rightarrow 2}\frac{\sqrt{9+x^2}+\sqrt{6-x}-2\sqrt{3-x}-2}{\sqrt{9+x^2}-1}\]

  10. amonoconnor
    • one year ago
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    That algebra is in fact wrong, correct, at least the 1st term of the numerator in the product?

  11. jim_thompson5910
    • one year ago
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    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

  12. jim_thompson5910
    • one year ago
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    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 \[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]

  13. amonoconnor
    • one year ago
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    Damn, alright...

  14. amonoconnor
    • one year ago
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    I'm going to take a minute to work through it if that's okay?

  15. jim_thompson5910
    • one year ago
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    notice this pairing here \[\Large \frac{\color{red}{\sqrt{6-x}-2}}{\sqrt{3-x}-1}\times\frac{\color{red}{\sqrt{6-x}+2}}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]

  16. jim_thompson5910
    • one year ago
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    and this pairing as well \[\Large \frac{\sqrt{6-x}-2}{\color{red}{\sqrt{3-x}-1}}\times\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\color{red}{\sqrt{3-x}+1}}\]

  17. amonoconnor
    • one year ago
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    Is this a rule for any situations with a radical in both the numerator and denominator?

  18. jim_thompson5910
    • one year ago
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    not that I can remember, but this trick of using 2 conjugates seems to work

  19. amonoconnor
    • one year ago
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    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)?

  20. jim_thompson5910
    • one year ago
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    \[\Large \sqrt{6-x}*\sqrt{3-x} = \sqrt{(6-x)*(3-x)}\] \[\Large \sqrt{6-x}*\sqrt{3-x} = \sqrt{18 - 9x + x^2}\]

  21. jim_thompson5910
    • one year ago
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    \[\Large (a+b)*(c+d) \ne a*c + b*d\]

  22. jim_thompson5910
    • one year ago
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    you have to use the FOIL rule

  23. amonoconnor
    • one year ago
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    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 ?

  24. jim_thompson5910
    • one year ago
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    ok go ahead

  25. amonoconnor
    • one year ago
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    \[\lim_{x \rightarrow 2}\frac{ \sqrt{6-x}-2 }{\sqrt{3-x}-1}*\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}*\frac{ \sqrt{3-x}+1 }{\sqrt{3-x}+1} = \lim_{x \rightarrow 2}\frac{(6-x)-4+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}{(3-x)-1+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}\]

  26. amonoconnor
    • one year ago
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    \[\lim_{x \rightarrow 2}\frac{(6-x)-4+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}{(3-x)-1+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}\]

  27. jim_thompson5910
    • one year ago
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    why do all that work? Seems way too tedious than it needs to be. Here's how I did it \[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\] \[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\] \[\Large \frac{6-x-4}{\sqrt{3-x}-1}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\] \[\Large \frac{6-x-4}{3-x-1}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\] \[\Large \frac{2-x}{2-x}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\] \[\Large \frac{1}{1}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\]

  28. amonoconnor
    • one year ago
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    Can I cancel out that string of common terms in the top and bottom?

  29. amonoconnor
    • one year ago
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    Sorry, in mine?

  30. jim_thompson5910
    • one year ago
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    `Can I cancel out that string of common terms in the top and bottom?` you mean like canceling the x terms in something like \[\Large \frac{x+3}{x+2}\]

  31. jim_thompson5910
    • one year ago
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    I'm not sure what you mean. Have a look at my steps. Let me know if they make sense or not

  32. amonoconnor
    • one year ago
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    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?

  33. amonoconnor
    • one year ago
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    That just makes the limit, or answer, go to zero then... that can't be right.

  34. jim_thompson5910
    • one year ago
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    no you can't do this \[\lim_{x \rightarrow 2}\frac{(6-x)-4+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}{(3-x)-1+\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}\] \[\lim_{x \rightarrow 2}\frac{(6-x)-4+\cancel{\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}}{(3-x)-1+\cancel{\sqrt{18-9x+x^2}+\sqrt{6-x}+2\sqrt{3-x}+2}}\]

  35. jim_thompson5910
    • one year ago
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    you agree with this step right? \[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]

  36. amonoconnor
    • one year ago
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    Yep, that's what I did:)

  37. jim_thompson5910
    • one year ago
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    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?

  38. amonoconnor
    • one year ago
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    Correct:) I think I got the multiplying across right, but I just don't know if you can break it down more :/

  39. jim_thompson5910
    • one year ago
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    ok so after doing that, we will get to this step \[\Large \frac{2-x}{\sqrt{3-x}-1}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]

  40. jim_thompson5910
    • one year ago
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    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 \[\Large \frac{2-x}{2-x}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\]

  41. amonoconnor
    • one year ago
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    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?

  42. jim_thompson5910
    • one year ago
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    I don't think there is a way to simplify that result you got, which is why I avoided expanding things out

  43. amonoconnor
    • one year ago
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    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...

  44. amonoconnor
    • one year ago
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    I don't get what to do then!

  45. jim_thompson5910
    • one year ago
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    do you see how I got to \[\Large \frac{2-x}{2-x}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\]

  46. amonoconnor
    • one year ago
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    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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is replying to Can someone tell me what button the professor is hitting...

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