anonymous
  • anonymous
Find lim (x^2-25)/(x^3-27) x-> 3 ? Please help?? (:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Luigi0210
  • Luigi0210
If I am correct, the limit DNE.
Luigi0210
  • Luigi0210
\[\lim_{x \rightarrow 3^-}f(x)=\infty \]\[\lim_{x \rightarrow 3^+}f(x)=- \infty \] \[\lim_{x \rightarrow 3^-} f(x) \neq \lim_{x \rightarrow 3^+} f(x)\]
Luigi0210
  • Luigi0210
|dw:1378701552558:dw|

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Luigi0210
  • Luigi0210
Does it make sense? @SoccerGirl13 :/
anonymous
  • anonymous
it's actually 2/15, but i don't know how to get that answer :(
anonymous
  • anonymous
and sory i wrote the equation wrong! DX
anonymous
  • anonymous
it's (x^2-9)/(x^3-27) x>3 ! D:
anonymous
  • anonymous
@Luigi0210
Luigi0210
  • Luigi0210
Oh, well factor the two equations, and what do you get?
anonymous
  • anonymous
(X-3) (X+3)
Luigi0210
  • Luigi0210
Now the denominator
anonymous
  • anonymous
(x+3)(x+3)(x-3)
Luigi0210
  • Luigi0210
You sure about that?
anonymous
  • anonymous
no D:
anonymous
  • anonymous
how do you factor it?
anonymous
  • anonymous
:(
Luigi0210
  • Luigi0210
\[a^3 - b^3 = (a - b) (a^2 + ab + b^2) \]
anonymous
  • anonymous
i'm lost ):
Luigi0210
  • Luigi0210
Okay, babysteps, do you know the cubic root of 27?
anonymous
  • anonymous
3
Luigi0210
  • Luigi0210
Yup, and you obviously know the cubic root of x^3? right?
anonymous
  • anonymous
27 :D
Luigi0210
  • Luigi0210
No, we're not plugging anything in, yet. I meant just the x^3
Luigi0210
  • Luigi0210
So the cubic root of 27 is 3 The cubic root of x^3 is x We have our a, x, and our b, 3.
anonymous
  • anonymous
uhuuu.. .__.
anonymous
  • anonymous
so once you figure that out then what? :o
Luigi0210
  • Luigi0210
Sorry about that delay, but you factor it you should end up with this: \[\lim_{x \rightarrow 3} \frac{(x-3)(x+3)}{(x-3)(x^2+3x+9)}\]
anonymous
  • anonymous
hm, okay. so then after that? :o
anonymous
  • anonymous
i honestly thought that you could just graph right away?
Luigi0210
  • Luigi0210
Oh, you wanted to graph it instead? Why didn't you just say so >.>
anonymous
  • anonymous
yeah! (: lol
anonymous
  • anonymous
i can graph it, i just don't know how to identify the limit of x>3 :(
Luigi0210
  • Luigi0210
Look for the value of y when x=3
Luigi0210
  • Luigi0210
This is why you need to solve it algebraically, because if you plug it in the way it is.. it will give you a hole at that point.
Luigi0210
  • Luigi0210
So if we continue with what we had, we can cancel out the (x-3) leaving us with: \[\lim_{x \rightarrow 3} \frac{x+3}{x^3+3x+9}\] Now plug in 3 and get: \[\frac{3+3}{3^2+3(3)+9}=\frac{6}{27}=\frac{2}{9}\]
Luigi0210
  • Luigi0210
Even wolfram says so: http://www.wolframalpha.com/input/?i=lim+%28x%5E2-9%29%2F%28x%5E3-27%29+x%3D%3E3+
anonymous
  • anonymous
ohhhhh, okay!
anonymous
  • anonymous
:D i see it
anonymous
  • anonymous
same for (x^2-25)/(x^3-125)
Luigi0210
  • Luigi0210
Yup, same process
anonymous
  • anonymous
(x+5)(x-5)/(x-5)(x^2+5x+25)
Luigi0210
  • Luigi0210
Mhm! What is the limit? Or is this just factoring practice?
anonymous
  • anonymous
x>5
Luigi0210
  • Luigi0210
Should of known, but yea you know what to do :P
anonymous
  • anonymous
2/15 ?(:
Luigi0210
  • Luigi0210
Bingo.
anonymous
  • anonymous
Mind helping me on another type of problem? :o (:
Luigi0210
  • Luigi0210
Open a new question. I have to leave soon :/
anonymous
  • anonymous
okay it'll be a quick one (:

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