Question

Does anyone know how to find if a hole as x is coming towards a value, and what h(x) will be at that point? \[h(x)= (x-4)/(2x^2-12x+16)\]

Answer 1

factor and cancel is the trick

Answer 2

this one factors as
\[\frac{x-4}{(x-4)(2x-4)}=\frac{1}{2x-4}\]\]

Answer 3

now if you want the limit as x -> 4 replace x by 4 to get your answer

Answer 4

I understand that 4 IS the hole, but the question is asking what h(x) is as x is coming towards \[2^{-}, 2^{+}, 4^{-}, 4^{+}\]

Answer 5

limit as x -> 4 from both directions is what you get if you replace x by 4

Answer 6

there is no limit as x ->2 since the denominator (of the factored thing) will have a 0 in it

Answer 7

so first two limits do not exist. this function has a vertical asymptote there. second two limits are 1/4

Answer 8

so 2^+ and 2^- h(x) is coming toward +/- oo?
That's true for 4 though, is it?

Answer 9

no the limit as x->4 is just 1/4

Answer 10

in other words, this function is really just
\[\frac{1}{2x-4}\] but not defined at 4

Answer 11

but clearly you can plug 4 in the factored one and if you do you get 1/4

Answer 12

but you may not plug 2 into the factored one. that is an asymptote and there is no limit

Answer 13

here is a nice picture. you can see that 4 is not a problem at all but 2 certainly is
http://www.wolframalpha.com/input/?i=%28x%E2%88%924%29%2F%282x^2%E2%88%9212x%2B16%29