# How do you find the lim (x^2-x+12) / (x+3) as x approaches -3?

Aug 4, 2015

${\lim}_{x \rightarrow - 3} \frac{{x}^{2} - x + 12}{x + 3}$ does not exist.

#### Explanation:

As $x \rightarrow - 3$,

we get the numerator: ${x}^{2} - x + 12 \rightarrow 9 + 3 + 12 = 24$

while the denominator: $x + 3 \rightarrow 0$

There is no number that the ratio is getting closer to.

In fact If $x$ is close to $- 3$ but a little left of $- 3$, then the denominator is a negative fraction very close to $0$, so the whole thing is a very big negative number.

We write: ${\lim}_{x \rightarrow - {3}^{-}} \frac{{x}^{2} - x + 12}{x + 3} = - \infty$ to indicate that the limit as $x$ approaches $- 3$ from the left does not exist because as $x$ approaches $- 3$ from the left, the ratio decreases without bound.
(Yes, that is an awful lot to write. That's why it's nice to have a good notation for it.)

A similar analysis on the right of $- 3$ will show us that limit as $x$ approaches $- 3$ from the right does not exist because as $x$ approaches $- 3$ from the right, the ratio increases without bound.

We write: ${\lim}_{x \rightarrow - {3}^{+}} \frac{{x}^{2} - x + 12}{x + 3} = \infty$

In my experience in the US, for the two sided limit, we write:

${\lim}_{x \rightarrow - 3} \frac{{x}^{2} - x + 12}{x + 3}$ does not exist.

or ${\lim}_{x \rightarrow - 3} \left\mid \frac{{x}^{2} - x + 12}{x + 3} \right\mid = \infty$

I have been told that the convention in other places is to write
${\lim}_{x \rightarrow - 3} \frac{{x}^{2} - x + 12}{x + 3} = \infty$ in this case.