Question

How do you evaluate `(x^2-16) /( x^2+3x-4)` as x approaches 1+?

Answer 1

Please see the explanation section, below.

Explanation:

First note that `lim_(xrarr1^+) (x^2-16) = -15`. (The limit of the numerator is not `0`.)

Furthermore `lim_(xrarr1^+) (x^2+3x-4) = 0` (The limit of the denominator is `0`.)

The form of this limit is `("non-"0)/0`, there is no limit (the limit does not exist).

But we can say more about why the limit does not exist. The one-sided limit does noty exist because the function is either increasing without bound or it is decreasing without bound as `xrarr1^+`. Let's figure out which.

The numerator is approaching a negative number while the denominator goes to `0` at `1`. This tells us that `x-1` is a faxctor of the deniminator.

`x^2+3x-4 = (x-1)(x+4)`

There is one factor approaching `0` and another approaching `5`. Since `xrarr1^+`, we have `x > 1`, so `x-1 > 0`. Therefore the denominator is approaching `0`, and is positive.

A negative numerator and a positive denominator give us negative values.

The quotient does not approach a limit because, as `x` approaches `1` from the right, the function decreases without bound.

We abbreviate the previous sentence by writing

`lim_(xrarr1^+) (x^2-16)/(x^2+3x-4) = -oo`.