Question

Factoring out \(x-3\) from \[\frac{\frac{1}{x-4}+1}{x-3}\].

Basically, I'm working on some limit questions, and I need to factor out \(x-3\) from the base so I can determine if there is a VA or not. The limit is \(x->3\), which makes both the denominator and numerator zero. Therefore, I need to get \(x-3\) into the numerator somehow so I can cancel it with the \(x-3\) in the denominator.

Answer 1

hey can you please type the question again it does not really make sense like that?

Answer 2

oh i just got it my screen was not showing it well

Answer 3

you have to first simplify the numerator

Answer 4

so that becomes (x+5)/(x-4)
and all of this is divided by x-3

Answer 5

Wow, that was dumb! I even multiplied by \(x-4\) top and bottom before, and didn't see, somehow, that 1+ x-4 in the numerator became x-3 (messy writing ftw!).

Answer 6

now you can write this as ((x+5)/(x-4)) * (x-3)

Answer 7

this ends up being:
\[\frac{ (x+5)(x-3) }{ (x-4) }\]
You cant simplify it anymore!

Answer 8

I mean there is nothing that can be cancelled out.

Answer 9

I think it becomes:
\[\frac{(x-3)}{(x-3)(x-4)}\]

Answer 10

i think you are multiplying x-3 to the denominator which is a common mistake

Answer 11

Because \[(\frac{1}{x-4}+1) * (x-4) = 1 + x-4\]

Answer 12

For what it's worth, it gave me the right answer!