Question

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!

Answer 1

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

Answer 2

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

Answer 3

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?

Answer 4

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

Answer 5

Is that what it should work out to? :/

Answer 6

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

Answer 7

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! :(

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

\[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]

Answer 13

Damn, alright...

Answer 14

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

Answer 15

notice this pairing here

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

Answer 16

and this pairing as well

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

Answer 17

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

Answer 18

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

Answer 19

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)?

Answer 20

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

Answer 21

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

Answer 22

you have to use the FOIL rule

Answer 23

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 ?

Answer 24

ok go ahead

Answer 25

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

Answer 26

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

Answer 27

why do all that work? Seems way too tedious than it needs to be. Here's how I did it

\[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\]
\[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]
\[\Large \frac{6-x-4}{\sqrt{3-x}-1}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]
\[\Large \frac{6-x-4}{3-x-1}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\]
\[\Large \frac{2-x}{2-x}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\]
\[\Large \frac{1}{1}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\]

Answer 28

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

Answer 29

Sorry, in mine?

Answer 30

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

you mean like canceling the x terms in something like

\[\Large \frac{x+3}{x+2}\]

Answer 31

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

Answer 32

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?

Answer 33

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

Answer 34

no you can't do this

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

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

Answer 35

you agree with this step right?

\[\Large \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\times\frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]

Answer 36

Yep, that's what I did:)

Answer 37

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?

Answer 38

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

Answer 39

ok so after doing that, we will get to this step

\[\Large \frac{2-x}{\sqrt{3-x}-1}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{\sqrt{3-x}+1}\]

Answer 40

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

\[\Large \frac{2-x}{2-x}\times\frac{1}{\sqrt{6-x}+2}\times\frac{\sqrt{3-x}+1}{1}\]

Answer 41

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?

Answer 42

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

Answer 43

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...

Answer 44

I don't get what to do then!

Answer 45

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

Answer 46

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!