Limits problem

1. Sep 17, 2008

Melawrghk

1. The problem statement, all variables and given/known data
Determine (i) limx->6- (46 + 41/(x-6))/(720 + 41/(x-6))

2. Relevant equations
None

3. The attempt at a solution
I simplified 41/(x-6) to be 4(x-6)-1, which then became ((1/4)x-6)-1
Then, that equals:
((1/4)x)-1
---------
((1/4)6)-1

Then if I substitute x=6, that becomes 1 and doesn't really help as I would get 47/721 for my limit. But I will also have to find the limits for that equation coming from the right (x->6+) and just (x->6). On the graph all those should be different, I think. Plus lim at x->6 shouldn't even exist. But I don't know. Help?

Last edited: Sep 17, 2008
2. Sep 17, 2008

Dick

Your exponent 'simplification' isn't correct. Just think about the problem first. If x->6- you should think of x as a number a little less than 6. Then 1/(x-6) is a negative number of large magnitude. If you take 4 to a power like that what happens? Say like 4^(-100)?

3. Sep 17, 2008

Melawrghk

Oh okay, so if I pick x=5.9999, then it will be 4^(-10000), making the 4^(1/(x-6)) negligible? In that case, limit would just be 46/720.
I get that. Now, if I'm looking on the other side of the "graph", if I were to pick a number like 6.001, then 4^(1/(x-6)) would become 4^1000 (or something), and the close I get to 6, the higher that number will be. Does that mean that I can neglect 46 and 720 because they're so small? In which case I'll have two same infinities dividing and that would result in 1. Is that correct?