Hint: rationalize the denominator. lim(x,y)→(0,0) for (−9x2+8y2) / ((−9x2+8y2+1)-1)

Question

anonymous · Answer

so many parentheses D:
If you know calculus you could just utilize L'Hopital's rule. Otherwise I suppose you'll just have to follow the hint.

turingtest · Answer

$$\lim_{(x,y) \rightarrow (0,0)}{-9x^2+8y^2 \over (-9x^2+8y^2+1)-1}$$right?

anonymous · Answer

I have to find the limit, and I am not sure how to rationalize the denom

anonymous · Answer

You have to multiply by the conjugate.

anonymous · Answer

To:TuringTest    Thats correct

turingtest · Answer

why can't we just get rid of that 1 in the denominator and say$$\lim_{(x,y) \rightarrow (0,0)}{-9x^2+8y^2\over-9x^2+8y^2}=1$$

turingtest · Answer

I don't see how to rationalize that denominator. It doesn't look irrational to me.

anonymous · Answer

Thats my problem
It think it may want polar cordinates, and idk how to do those either

anonymous · Answer

if we do your method, wouldnt it just turn into  1/(1-1) making it irrational

turingtest · Answer

How did you get a 1 in the numerator?
as far as the denominator, look at what I did above. You can ignore those parentheses, they are meaningless. So if you rewrite the denominator as$$-9x^2+8y^2+1-1=-9x^2+8y^2$$you see it's the same as the numerator. You clearly don't need to plug in (0,0) when the top and bottom cancel to give 1.

anonymous · Answer

Thats wrong
I submitted 1 and it came up  incorrect
Im an idiot, I type the problem wrong!  The part in parenthese is sqrt's that is why the neg 1 is on the outside

turingtest · Answer

Ahhh, that explains everything. It also makes the problem much more difficult. Let me see now...$$\lim_{(x,y) \rightarrow (0,0)}={-9x^2+8x^2\over \sqrt{-9x^2+8y^2+1}-1}$$right? Hmm... let me try some things on paper, it'll be faster that way. I'll have to get back to you (hopefully not too long).

anonymous · Answer

Im so sorry for the confusion!

turingtest · Answer

it's all good... 
working on it...

turingtest · Answer

ok it's not so bad, check it out... after multiplying top and bottom by the conjugate we get$${(-9x^2+8x^2)(\sqrt{-9x^2+8y^2+1}+1)\over (-9x^2+8x^2)}=\sqrt{-9x^2+8y^2+1}+1$$so$$\lim_{(x,y) \rightarrow (0,0)}\sqrt{-9x^2+8y^2+1}+1=2$$I hope you see how I got the denominator after multiplying by the conjugate.

anonymous · Answer

Thats correct, wow good job

anonymous · Answer

I hate conjugates, but yeah I can see what you did

turingtest · Answer

Thanks, it was kinda tough. I don't usually beg but... medal please? I took a bit of time on this one and I think I earned it.

turingtest · Answer

thx

anonymous · Answer

dude yeah!!   I give medals to all who help and show effort
I can most more similar problems if you want more medals

anonymous · Answer

none will be as hard im sure, and i dont need work

anonymous · Answer

i just need the answer haha

turingtest · Answer

I'm getting drowsy, going to sleep soon. However most tutors here can't do these so if I can't do it tonight I'll look up your questions tomorrow. If you think they're easier though and you just need answers you can try me. Just post separately.

anonymous · Answer

How do i post separately, can I direct them to you some how?

turingtest · Answer

Not that I know of, but I watch for the most recent questions. Just post it normally and I should find it.

anonymous · Answer

ok thanks!

turingtest · Answer

oh, and if you become my fan I can find your questions more easily when you are online, so you may want to do that.