## anonymous 5 years ago Evaluate limit (x+1)/(2x-6)/(x^2-4) as x approaches 2

1. nowhereman

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

2. anonymous

sorry

3. anonymous

its 3 things dividing each other

4. anonymous

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

5. anonymous

is that better?

6. anonymous

its an improper fraction.

7. nowhereman

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

8. anonymous

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

9. 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.

10. anonymous

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

11. nowhereman

Then the formula must be different.

12. anonymous

I simplified it a bit.

13. anonymous

The original formula is...

14. anonymous

$((x/2)+(1/(x-3))/(x^2-4)$

15. nowhereman

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

16. anonymous

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

17. anonymous

That's the original expression.

18. anonymous

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

19. anonymous

and remove the discontinuity

20. nowhereman

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

21. anonymous

perfect!

22. anonymous

Thank you.

23. anonymous

wait...

24. anonymous

nevermind.