Question

for the function $f(x)$ shown below, determine $lim_{x
ightarrow - 1}f(x)$. $f(x)=\begin{cases}-2x + 2& \text{for }x > - 1\\3 + x^{2}&\text{for }x < - 1end{cases}$ answer attempt 1 out of 2

Explanation:

Step1: Find left - hand limit

We find $\lim_{x
ightarrow - 1^{-}}f(x)$. Since $x
ightarrow - 1^{-}$ means $x < - 1$, we use $f(x)=3 + x^{2}$.
$\lim_{x
ightarrow - 1^{-}}f(x)=\lim_{x
ightarrow - 1^{-}}(3 + x^{2})=3+( - 1)^{2}=3 + 1=4$

Step2: Find right - hand limit

We find $\lim_{x
ightarrow - 1^{+}}f(x)$. Since $x
ightarrow - 1^{+}$ means $x > - 1$, we use $f(x)=-2x + 2$.
$\lim_{x
ightarrow - 1^{+}}f(x)=\lim_{x
ightarrow - 1^{+}}(-2x + 2)=-2\times(-1)+2=2 + 2=4$

Step3: Determine the limit

Since $\lim_{x
ightarrow - 1^{-}}f(x)=\lim_{x
ightarrow - 1^{+}}f(x)=4$, then $\lim_{x
ightarrow - 1}f(x)=4$.

Answer:

$4$