Question

a function is said to have a vertical asymptote wherever the limit on the left or right (or both) is either positive or negative infinity. for example, the function $f(x)=\frac{-3(x + 2)}{x^{2}+4x + 4}$ has a vertical asymptote at $x=-2$. find each of the following limits. $lim_{x
ightarrow - 2^{-}}\frac{-3(x + 2)}{x^{2}+4x + 4}=square$ help (limits) $lim_{x
ightarrow - 2^{+}}\frac{-3(x + 2)}{x^{2}+4x + 4}=square$ help (limits) $lim_{x
ightarrow - 2}\frac{-3(x + 2)}{x^{2}+4x + 4}=square$ help (limits)

Explanation:

Step1: Factor the denominator

First, factor $x^{2}+4x + 4=(x + 2)^{2}$. So the function becomes $f(x)=\frac{-3(x + 2)}{(x + 2)^{2}}=\frac{-3}{x + 2},x
eq - 2$.

Step2: Find $\lim_{x

ightarrow - 2^{-}}\frac{-3(x + 2)}{x^{2}+4x + 4}$
As $x
ightarrow - 2^{-}$, let $x=-2 - h$, where $h>0$ and $h
ightarrow0$. Then $\frac{-3}{x + 2}=\frac{-3}{-2 - h+2}= \frac{3}{h}$. As $h
ightarrow0^{+}$, $\lim_{x
ightarrow - 2^{-}}\frac{-3(x + 2)}{x^{2}+4x + 4}=\lim_{h
ightarrow0^{+}}\frac{3}{h}=+\infty$.

Step3: Find $\lim_{x

ightarrow - 2^{+}}\frac{-3(x + 2)}{x^{2}+4x + 4}$
As $x
ightarrow - 2^{+}$, let $x=-2 + h$, where $h>0$ and $h
ightarrow0$. Then $\frac{-3}{x + 2}=\frac{-3}{-2 + h+2}=-\frac{3}{h}$. As $h
ightarrow0^{+}$, $\lim_{x
ightarrow - 2^{+}}\frac{-3(x + 2)}{x^{2}+4x + 4}=\lim_{h
ightarrow0^{+}}-\frac{3}{h}=-\infty$.

Step4: Find $\lim_{x

ightarrow - 2}\frac{-3(x + 2)}{x^{2}+4x + 4}$
Since $\lim_{x
ightarrow - 2^{-}}\frac{-3(x + 2)}{x^{2}+4x + 4}=+\infty$ and $\lim_{x
ightarrow - 2^{+}}\frac{-3(x + 2)}{x^{2}+4x + 4}=-\infty$, the two - sided limit $\lim_{x
ightarrow - 2}\frac{-3(x + 2)}{x^{2}+4x + 4}$ does not exist.

Answer:

$\lim_{x
ightarrow - 2^{-}}\frac{-3(x + 2)}{x^{2}+4x + 4}=+\infty$
$\lim_{x
ightarrow - 2^{+}}\frac{-3(x + 2)}{x^{2}+4x + 4}=-\infty$
$\lim_{x
ightarrow - 2}\frac{-3(x + 2)}{x^{2}+4x + 4}$ does not exist.