Question

How do you find the limit of `(x^2 - 8) / (8x-16)` as x approaches `2^+`?

Answer 1

`lim_(xrarr2^+)(x^2-8)/(8x-16) = color(green)(-oo)`

Explanation:

Given:
`color(white)("XXX")lim_(xrarr2^+)(x^2-8)/(8x-16)`

If `xrarr2^+` then `x > 2`
and therefore
`color(white)("XXX")(8x-16) > 0`
(i.e. `8x-16` is positive as `xrarr2^+`)

At `x=2` (and thus as `xrarr2^+`)
`color(white)("XXX")x^2-8=-4`

Therefore
`color(white)("XXX")lim_(xrarr2^+) (x^2-8)/(8x-16)`

`color(white)("XXX")="non-zero, negative value"/("positive value approaching " 0)=-oo`

We could also see this by graphing the given expression: