Limiting the Equation: Determine (i) limx->6-

[Melawrghk]

Homework Statement

Determine (i) lim_x->6^- (46 + 4^1/(x-6))/(720 + 4^1/(x-6))

Homework Equations

None

The Attempt at a Solution

I simplified 4^1/(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?

 

[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)?

 

[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?

Thank you for your help.

 

[Dick]

That's correct.