Question

Evaluate limit (x+1)/(2x-6)/(x^2-4) as x approaches 2

Answer 1

You have to place more parentheses, to give that term a meaning!

Answer 2

sorry

Answer 3

its 3 things dividing each other

Answer 4

so on top is \[((x+1)\div(2x-6))\div(x^2-4)\]

Answer 5

is that better?

Answer 6

its an improper fraction.

Answer 7

Yes. The limit does not exist, there is a pole at x=2.

Answer 8

Right, there should be a removable discontinuity because the answer is -1/8

Answer 9

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.

Answer 10

So, the answer does not exist? Weird, because the answer key says that its -1/8.

Answer 11

Then the formula must be different.

Answer 12

I simplified it a bit.

Answer 13

The original formula is...

Answer 14

\[((x/2)+(1/(x-3))/(x^2-4)\]

Answer 15

Then you must have made a mistake, because here the numerator tends to 0 too.

Answer 16

I combined the numerator to find a common denominator for them.

Answer 17

That's the original expression.

Answer 18

I'm assuming I have to manipulate it to cancel out x-2 or x+2.

Answer 19

and remove the discontinuity

Answer 20

Yes, \(x-2\). But it should work: \[\frac x 2 + \frac 1 {x-3} = \frac{x(x-3) + 2}{2(x-3)} = \frac{x^2-3x+2}{2(x-3)} = \frac{(x-2)(x-1)}{2(x-3)}\]

Answer 21

perfect!

Answer 22

Thank you.

Answer 23

wait...

Answer 24

nevermind.