Question

for the function $f(x)$ shown below, determine $lim_{x
ightarrow1}f(x)$. $f(x)=\begin{cases}-2x^{2}-3&\text{for }xleq1\\-4 + 4x&\text{for }x>1end{cases}$ answer attempt 1 out of 2

Explanation:

Step1: Find left - hand limit

For $x\leq1$, $f(x)=-2x^{2}-3$. Calculate $\lim_{x
ightarrow1^{-}}f(x)$:
Substitute $x = 1$ into $-2x^{2}-3$. So, $\lim_{x
ightarrow1^{-}}f(x)=-2(1)^{2}-3=-2 - 3=-5$.

Step2: Find right - hand limit

For $x>1$, $f(x)=-4 + 4x$. Calculate $\lim_{x
ightarrow1^{+}}f(x)$:
Substitute $x = 1$ into $-4 + 4x$. So, $\lim_{x
ightarrow1^{+}}f(x)=-4+4(1)=0$.

Step3: Check if limit exists

Since $\lim_{x
ightarrow1^{-}}f(x)=-5$ and $\lim_{x
ightarrow1^{+}}f(x)=0$, and $\lim_{x
ightarrow1^{-}}f(x)
eq\lim_{x
ightarrow1^{+}}f(x)$.

Answer:

DNE