Question

for the function $f(x)$ shown below, determine $lim_{x
ightarrow2}f(x)$. $f(x)=\begin{cases}-4x + 3& \text{for }xgeq2\\2x^{2}-10& \text{for }x < 2end{cases}$ answer attempt 1 out of 2 dne

Explanation:

Step1: Find left - hand limit

We use the part of the function for $x < 2$. So, $\lim_{x
ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{-}}(2x^{2}-10)$. Substitute $x = 2$ into $2x^{2}-10$: $2\times2^{2}-10=2\times4 - 10=8 - 10=-2$.

Step2: Find right - hand limit

We use the part of the function for $x\geq2$. So, $\lim_{x
ightarrow2^{+}}f(x)=\lim_{x
ightarrow2^{+}}(-4x + 3)$. Substitute $x = 2$ into $-4x + 3$: $-4\times2+3=-8 + 3=-5$.

Step3: Check if limit exists

Since $\lim_{x
ightarrow2^{-}}f(x)=-2$ and $\lim_{x
ightarrow2^{+}}f(x)=-5$, and $\lim_{x
ightarrow2^{-}}f(x)
eq\lim_{x
ightarrow2^{+}}f(x)$, the limit $\lim_{x
ightarrow2}f(x)$ does not exist.

Answer:

DNE