Question

exercises 2.3 the limit laws score: 15/21 answered: 15/21 question 16 let $f(x)=\begin{cases}10x + 10&\text{if }x<10\cx&\text{if }x > 10end{cases}$ $lim_{x
ightarrow10^{-}}f(x)=$ $lim_{x
ightarrow10^{+}}f(x)=$ if $lim_{x
ightarrow10}f(x)$ exists, then it must be (enter a numerical value). this means $c=$ question help: video written example message instructor

Explanation:

Step1: Find left - hand limit

We use the part of the function for $x < 10$. So, $\lim_{x
ightarrow10^{-}}f(x)=\lim_{x
ightarrow10^{-}}(10x + 10)$. Substitute $x = 10$ into $10x+10$: $10\times10 + 10=110$.

Step2: Find right - hand limit

We use the part of the function for $x>10$. So, $\lim_{x
ightarrow10^{+}}f(x)=\lim_{x
ightarrow10^{+}}(cx)=10c$.

Step3: Determine the limit value if it exists

For $\lim_{x
ightarrow10}f(x)$ to exist, $\lim_{x
ightarrow10^{-}}f(x)=\lim_{x
ightarrow10^{+}}f(x)$. Since $\lim_{x
ightarrow10^{-}}f(x) = 110$ and $\lim_{x
ightarrow10^{+}}f(x)=10c$, when the limit exists, $\lim_{x
ightarrow10}f(x)=110$ (because the left - hand and right - hand limits must be equal).

Step4: Solve for c

Set $10c=110$. Divide both sides by 10: $c=\frac{110}{10}=11$.

Answer:

$\lim_{x
ightarrow10^{-}}f(x)=110$
$\lim_{x
ightarrow10^{+}}f(x)=110$
If $\lim_{x
ightarrow10}f(x)$ exists, then it must be $110$
$c = 11$