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anonymous
 5 years ago
Evaluate limit (x+1)/(2x6)/(x^24) as x approaches 2
anonymous
 5 years ago
Evaluate limit (x+1)/(2x6)/(x^24) as x approaches 2

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nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.1You have to place more parentheses, to give that term a meaning!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its 3 things dividing each other

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so on top is \[((x+1)\div(2x6))\div(x^24)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its an improper fraction.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.1Yes. The limit does not exist, there is a pole at x=2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Right, there should be a removable discontinuity because the answer is 1/8

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.1No, \(x^2  4\) tends to 0 and as it is in the denominator, plus the numerator is continuous at x=2, there must be a pole.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, the answer does not exist? Weird, because the answer key says that its 1/8.

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.1Then the formula must be different.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I simplified it a bit.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The original formula is...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[((x/2)+(1/(x3))/(x^24)\]

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.1Then you must have made a mistake, because here the numerator tends to 0 too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I combined the numerator to find a common denominator for them.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's the original expression.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm assuming I have to manipulate it to cancel out x2 or x+2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and remove the discontinuity

nowhereman
 5 years ago
Best ResponseYou've already chosen the best response.1Yes, \(x2\). But it should work: \[\frac x 2 + \frac 1 {x3} = \frac{x(x3) + 2}{2(x3)} = \frac{x^23x+2}{2(x3)} = \frac{(x2)(x1)}{2(x3)}\]
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