anonymous
  • anonymous
Evaluate limit (x+1)/(2x-6)/(x^2-4) as x approaches 2
Mathematics
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anonymous
  • anonymous
Evaluate limit (x+1)/(2x-6)/(x^2-4) as x approaches 2
Mathematics
jamiebookeater
  • jamiebookeater
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nowhereman
  • nowhereman
You have to place more parentheses, to give that term a meaning!
anonymous
  • anonymous
sorry
anonymous
  • anonymous
its 3 things dividing each other

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anonymous
  • anonymous
so on top is \[((x+1)\div(2x-6))\div(x^2-4)\]
anonymous
  • anonymous
is that better?
anonymous
  • anonymous
its an improper fraction.
nowhereman
  • nowhereman
Yes. The limit does not exist, there is a pole at x=2.
anonymous
  • anonymous
Right, there should be a removable discontinuity because the answer is -1/8
nowhereman
  • nowhereman
No, \(x^2 - 4\) tends to 0 and as it is in the denominator, plus the numerator is continuous at x=2, there must be a pole.
anonymous
  • anonymous
So, the answer does not exist? Weird, because the answer key says that its -1/8.
nowhereman
  • nowhereman
Then the formula must be different.
anonymous
  • anonymous
I simplified it a bit.
anonymous
  • anonymous
The original formula is...
anonymous
  • anonymous
\[((x/2)+(1/(x-3))/(x^2-4)\]
nowhereman
  • nowhereman
Then you must have made a mistake, because here the numerator tends to 0 too.
anonymous
  • anonymous
I combined the numerator to find a common denominator for them.
anonymous
  • anonymous
That's the original expression.
anonymous
  • anonymous
I'm assuming I have to manipulate it to cancel out x-2 or x+2.
anonymous
  • anonymous
and remove the discontinuity
nowhereman
  • nowhereman
Yes, \(x-2\). But it should work: \[\frac x 2 + \frac 1 {x-3} = \frac{x(x-3) + 2}{2(x-3)} = \frac{x^2-3x+2}{2(x-3)} = \frac{(x-2)(x-1)}{2(x-3)}\]
anonymous
  • anonymous
perfect!
anonymous
  • anonymous
Thank you.
anonymous
  • anonymous
wait...
anonymous
  • anonymous
nevermind.

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