Question

for the function $f(x)$ shown below, determine $lim_{x
ightarrow - 1}f(x)$. $f(x)=\begin{cases}-2x^{2}+4& \text{for }xleq - 1\\3x + 5& \text{for }x>-1end{cases}$ 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, $\lim_{x
ightarrow - 1^{-}}f(x)=\lim_{x
ightarrow - 1^{-}}(-2x^{2}+4)$. Substitute $x = - 1$ into $-2x^{2}+4$: $-2(-1)^{2}+4=-2 + 4=2$.

Step2: Calculate right - hand limit

We use the part of the function for $x>-1$. So, $\lim_{x
ightarrow - 1^{+}}f(x)=\lim_{x
ightarrow - 1^{+}}(3x + 5)$. Substitute $x=-1$ into $3x + 5$: $3(-1)+5=-3 + 5=2$.

Step3: Determine the limit

Since $\lim_{x
ightarrow - 1^{-}}f(x)=\lim_{x
ightarrow - 1^{+}}f(x)=2$, then $\lim_{x
ightarrow - 1}f(x)=2$.

Answer:

$2$