Question

let $f(x)=\begin{cases}-\frac{4}{x + 2},& \text{if }x < - 2\\4x + 11,& \text{if }x > - 2end{cases}$. calculate the following limits. $lim_{x
ightarrow - 2^{-}}f(x)=$ $lim_{x
ightarrow - 2^{+}}f(x)=$ $lim_{x
ightarrow - 2}f(x)=$

Explanation:

Step1: Find left - hand limit

We use the part of the function for $x < - 2$. So, $\lim_{x
ightarrow - 2^{-}}f(x)=\lim_{x
ightarrow - 2^{-}}(-\frac{4}{x + 2})$. Substitute $x=-2 - h$ where $h
ightarrow0^{+}$. Then $\lim_{h
ightarrow0^{+}}(-\frac{4}{-2 - h+2})=\lim_{h
ightarrow0^{+}}(\frac{4}{h})=\infty$.

Step2: Find right - hand limit

We use the part of the function for $x > - 2$. So, $\lim_{x
ightarrow - 2^{+}}f(x)=\lim_{x
ightarrow - 2^{+}}(4x + 11)$. Substitute $x=-2$ into $4x + 11$. Then $4\times(-2)+11=-8 + 11 = 3$.

Step3: Determine the two - sided limit

Since $\lim_{x
ightarrow - 2^{-}}f(x)
eq\lim_{x
ightarrow - 2^{+}}f(x)$, $\lim_{x
ightarrow - 2}f(x)$ does not exist.

Answer:

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