Question

let $f(x)=\begin{cases}-x - x^{2}&\text{if }xleq3\\4x - 8&\text{if }x > 3end{cases}$. calculate the following limits. enter \dne\ if the limit does not exist. $lim_{x
ightarrow3^{-}}f(x)=$ $lim_{x
ightarrow3^{+}}f(x)=$ $lim_{x
ightarrow3}f(x)=$ question help: video written example message instructor

Explanation:

Step1: Find left - hand limit

For $\lim_{x
ightarrow3^{-}}f(x)$, since $x
ightarrow3^{-}$ means $x < 3$, we use $f(x)=-x - x^{2}$.
Substitute $x = 3$ into $-x - x^{2}$: $-(3)-(3)^{2}=-3 - 9=-12$.

Step2: Find right - hand limit

For $\lim_{x
ightarrow3^{+}}f(x)$, since $x
ightarrow3^{+}$ means $x>3$, we use $f(x)=4x - 8$.
Substitute $x = 3$ into $4x - 8$: $4\times3-8=12 - 8 = 4$.

Step3: Find overall limit

The overall limit $\lim_{x
ightarrow3}f(x)$ exists if and only if $\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{+}}f(x)$.
Since $\lim_{x
ightarrow3^{-}}f(x)=-12$ and $\lim_{x
ightarrow3^{+}}f(x)=4$, $\lim_{x
ightarrow3}f(x)$ does not exist (DNE).

Answer:

$\lim_{x
ightarrow3^{-}}f(x)=-12$
$\lim_{x
ightarrow3^{+}}f(x)=4$
$\lim_{x
ightarrow3}f(x)=DNE$