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

Jun 10, 2016

Please see the explanation section, below.

#### Explanation:

First note that ${\lim}_{x \rightarrow {1}^{+}} \left({x}^{2} - 16\right) = - 15$. (The limit of the numerator is not $0$.)

Furthermore ${\lim}_{x \rightarrow {1}^{+}} \left({x}^{2} + 3 x - 4\right) = 0$ (The limit of the denominator is $0$.)

The form of this limit is $\frac{\text{non-} 0}{0}$, 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 $x \rightarrow {1}^{+}$. 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.

${x}^{2} + 3 x - 4 = \left(x - 1\right) \left(x + 4\right)$

There is one factor approaching $0$ and another approaching $5$. Since $x \rightarrow {1}^{+}$, 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

${\lim}_{x \rightarrow {1}^{+}} \frac{{x}^{2} - 16}{{x}^{2} + 3 x - 4} = - \infty$.