Question

Evaluate the limit using squeeze theorem:
limit as x approaches 4 for (x^2-16)(x-4/|x-4|)

Answer 1

So you have:\[\lim_{x \rightarrow 4}(x^2-16)\frac{ x-4 }{ |x-4| }\]
Is that right?

Answer 2

yes

Answer 3

i figured i could squeeze the |x-4| between -|x-4| and |x-4| giving me and answer of 0 for both and then a 0 for the denominator by the squeeze theorem. I can then split the functions apart and when I plug in 0 for x they all equal 0, and 0*0-0=0. Can i do that? I used the quotient law to bring the zero out from underneath the fraction. I know the answer is 0 i just have to show proper work

Answer 4

Or can I not use the quotient theorem because the limit for |x-4| in the denominator doesn't technically exist?

Answer 5

OK. You've done quite a bit already!
Before applying any theorems, I would try to simplifty the formula a bit.
I hate these |absolute values| in a formula, so I always try to replace them:
|x-4|=x-4 if x>4
|x-4|=4-x if x<4
x=4 is not allowed, so we do not bother with it.

Answer 6

If I expand x²-16 and split up the limit, I get:\[\lim_{x \rightarrow 4^+}\frac{ (x+4)(x-4)^2 }{ x-4 }=\lim_{x \rightarrow 4^+}(x+4)(x-4)=0\]And\[\lim_{x \rightarrow 4^-}\frac{ (x+4)(x-4)^2 }{ 4-x }=-\lim_{x \rightarrow 4^-}\frac{ (x+4)(x-4)^2 }{ x-4 }=-\lim_{x \rightarrow 4^+}(x+4)(x-4)=0\]
So both left and right hand limits are equal (0), and therefore the limit is 0.
Unforunately, I never used any additional theorem... :(

Answer 7

Well your work makes more sense than mine, so thanks! I think it was really just kind of a challenge problem that I don't have to worry about popping up again. If it does, itll be on a test and I can do whatever work I want. Ill do what you did! Thanks so much for the thorough reply!

Answer 8

Welcome!