How do you find the limit of \lim _ { x \rightarrow - 2} \frac { x + 2} { x ^ { 2} + 6x + 8}?

Aug 30, 2017

${\lim}_{x \rightarrow - 2} \frac{x + 2}{{x}^{2} + 6 x + 8} = \frac{1}{2}$

Explanation:

Notice that plugging in $x = - 2$ in the function's current form gives $\frac{- 2 + 2}{{\left(- 2\right)}^{2} + 6 \left(- 2\right) + 8} = \frac{0}{0}$, so the limit is currently in an indeterminate form and we cant yet determine the limit.

However, notice that we can factor the denominator of the fraction:

${\lim}_{x \rightarrow - 2} \frac{x + 2}{{x}^{2} + 6 x + 8} = {\lim}_{x \rightarrow - 2} \frac{x + 2}{\left(x + 2\right) \left(x + 4\right)}$

Now we can see why plugging in $x = - 2$ into the denominator creates a $0$. However, we can cancel the $\left(x + 2\right)$ terms in the numerator and denominator:

$= {\lim}_{x \rightarrow - 2} \frac{1}{x + 4}$

In other words, the function $\frac{x + 2}{{x}^{2} + 6 x + 8}$ is identical to the function $\frac{1}{x + 4}$ except at the point $x = - 2$. The former function is identical to the latter function except that the first function has a hole at $x = - 2$. So, the limit is where the function should lie, which can be found by plugging $x = - 2$ into the simplified function, $\frac{1}{x + 4}$.

$= \frac{1}{- 2 + 4} = \frac{1}{2}$