checking work on finding limit

Question

anonymous · Answer

this is what I have so far it is x approaches -3 from the left
$$\frac{ x }{ \sqrt{x ^{2}-9} }$$
multiply by the radical to get it out of the denominator
$$\frac{ x \sqrt{x ^{2}-9} }{ x ^{2}-9 }$$
factor and that is where I am
$$\frac{ x \sqrt{x ^{2}-9} }{ (x+3)(x-3) }$$

anonymous · Answer

I was thinking now it would be better multiplying by the conjugate radical, but I am not sure how that would turn out on bottom

anonymous · Answer

so like $$\sqrt{x ^{2}+9}$$

experimentx · Answer

did you put this thing on wolfram?

anonymous · Answer

no but I uses a graphing utility and it approaches negative infinity

anonymous · Answer

so is the answer just negative infinity/ does not exist

anonymous · Answer

I just mainly want to know if the work I did was right or if there is a correction

experimentx · Answer

you don't get negative infinity ... you get complex infinity. that it is.

anonymous · Answer

I dont know what complex infinity is

experimentx · Answer

that is also a kind of infinty. but in complex number system.

debbieg · Answer

Wolfram says -infinty from the left, complex infin from the right?

debbieg · Answer

I suspect that's why he was only asked to find the limit approaching from the left...

anonymous · Answer

I just want to know if my math work at the top is correct or if I made some mistake or if there is more simplifying I can do

debbieg · Answer

It looks right to me.

experimentx · Answer

no need to do that. just put x->3-

anonymous · Answer

alright thanks I just wanted I double check I appreciate your help guys

jdoe0001 · Answer

yeah... factoring wise, is fine, but as experimentX  said, no  need for any simpifying or factoring, just keeping in mind that, x < -3, wil not give you an undefined fraction

debbieg · Answer

Right.... you don't need to simplify. You can put in x=-3.

Num'r <0, den'r=0, so limit->-infinity

debbieg · Answer

Limit of quotient where num'r is non-zero and den'r is 0 = +/- infinity.

You can just "plug in" x=-3 in all of these expressions, because you have either a polynomial, or the sq root of a polynomial.

experimentx · Answer

x->3- means x^2<9
this should be square root of negative which is complex infinity rather than -infinity

experimentx · Answer

check here http://www.wolframalpha.com/input/?i=limit+x-%3E3+x%2Fsqrt%28x%5E2+-+9%29