Question

What is `lim_(->-oo) f(x) = absx/(16-x^2)`?

Answer 1

`lim_(x rarr -oo)|x|/(16-x^2) = 0`

Explanation:

Since we're dealing with `-oo` we can say `|x| = -x`, so

`lim_(x rarr -oo)|x|/(16-x^2) = lim_(x rarr -oo)-x/(16-x^2)`

Now, with that out of the way, factor out an `x` on the denominator
`lim_(x rarr -oo)-x/(16-x^2) = lim_(x rarr -oo)x/(x(16/x - x))`

Cancel the two `x`s otu

`lim_(x rarr -oo) = 1/(16/x - x)`

Now, because that `16/x` will become a `0` in the limit, which will make the limit as a whole become a `0`, or

`lim_(x rarr -oo) 1/(0 - x) = lim_(x rarr -oo) 1/x = 0`