a function is said to have a vertical asymptote wherever the limit on the left or right (or both) is either…

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. $lim_{x\rightarrow - 2^{-}}\frac{-3(x + 2)}{x^{2}+4x + 4}=square$ help (limits) $lim_{x\rightarrow - 2^{+}}\frac{-3(x + 2)}{x^{2}+4x + 4}=square$ help (limits) $lim_{x\rightarrow - 2}\frac{-3(x + 2)}{x^{2}+4x + 4}=square$ help (limits)

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. $lim_{x\rightarrow - 2^{-}}\frac{-3(x + 2)}{x^{2}+4x + 4}=square$ help (limits) $lim_{x\rightarrow - 2^{+}}\frac{-3(x + 2)}{x^{2}+4x + 4}=square$ help (limits) $lim_{x\rightarrow - 2}\frac{-3(x + 2)}{x^{2}+4x + 4}=square$ help (limits)

Answer

Explanation:

Step1: Factor the denominator

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

Step2: Find $\lim_{x\rightarrow - 2^{-}}\frac{-3(x + 2)}{x^{2}+4x + 4}$

As $x\rightarrow - 2^{-}$, let $x=-2 - h$, where $h>0$ and $h\rightarrow0$. Then $\frac{-3}{x + 2}=\frac{-3}{-2 - h+2}= \frac{3}{h}$. As $h\rightarrow0^{+}$, $\lim_{x\rightarrow - 2^{-}}\frac{-3(x + 2)}{x^{2}+4x + 4}=\lim_{h\rightarrow0^{+}}\frac{3}{h}=+\infty$.

Step3: Find $\lim_{x\rightarrow - 2^{+}}\frac{-3(x + 2)}{x^{2}+4x + 4}$

As $x\rightarrow - 2^{+}$, let $x=-2 + h$, where $h>0$ and $h\rightarrow0$. Then $\frac{-3}{x + 2}=\frac{-3}{-2 + h+2}=-\frac{3}{h}$. As $h\rightarrow0^{+}$, $\lim_{x\rightarrow - 2^{+}}\frac{-3(x + 2)}{x^{2}+4x + 4}=\lim_{h\rightarrow0^{+}}-\frac{3}{h}=-\infty$.

Step4: Find $\lim_{x\rightarrow - 2}\frac{-3(x + 2)}{x^{2}+4x + 4}$

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

Answer:

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