Question

for the function $f(x)$ shown below, determine $lim_{x
ightarrow - 1}f(x)$. $f(x)=\begin{cases}2x^{2}-8& \text{for }x > - 1\\4 - 2x& \text{for }xleq - 1end{cases}$ answer attempt 1 out of 2 dne

Explanation:

Step1: Calculate left - hand limit

We use the part of the function for $x\leq - 1$. So, $\lim_{x
ightarrow - 1^{-}}f(x)=\lim_{x
ightarrow - 1^{-}}(4 - 2x)$. Substitute $x=-1$ into $4 - 2x$, we get $4-2\times(-1)=4 + 2=6$.

Step2: Calculate right - hand limit

We use the part of the function for $x > - 1$. So, $\lim_{x
ightarrow - 1^{+}}f(x)=\lim_{x
ightarrow - 1^{+}}(2x^{2}-8)$. Substitute $x = - 1$ into $2x^{2}-8$, we get $2\times(-1)^{2}-8=2 - 8=-6$.

Step3: Check if the limits are equal

Since $\lim_{x
ightarrow - 1^{-}}f(x)=6$ and $\lim_{x
ightarrow - 1^{+}}f(x)=-6$, and $6
eq - 6$.

Answer:

DNE