Question

hw5 the limit laws (target l4; §2.3)
score: 11/13 answered: 12/13
question 13
let $f(x)=\begin{cases}-3x + 5&\text{if }xleq1\\-6x + 6&\text{if }x>1end{cases}$
find $lim_{x
ightarrow1}f(x)$, if it exists.
2
2
2
dne
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Explanation:

Step1: Find left - hand limit

We use the part of the piece - wise function for $x\leq1$.
$\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{-}}(-3x + 5)$
Substitute $x = 1$ into $-3x + 5$:
$-3(1)+5=2$

Step2: Find right - hand limit

We use the part of the piece - wise function for $x>1$.
$\lim_{x
ightarrow1^{+}}f(x)=\lim_{x
ightarrow1^{+}}(-6x + 6)$
Substitute $x = 1$ into $-6x + 6$:
$-6(1)+6=0$

Step3: Check if limit exists

Since $\lim_{x
ightarrow1^{-}}f(x)=2$ and $\lim_{x
ightarrow1^{+}}f(x)=0$, and $\lim_{x
ightarrow1^{-}}f(x)
eq\lim_{x
ightarrow1^{+}}f(x)$.
The limit $\lim_{x
ightarrow1}f(x)$ does not exist.

Answer:

DNE