Question

determine the value of c so that $lim_{x
ightarrow5}f(x)$ exists: $f(x)=\begin{cases}\frac{1}{5}x + c, &\text{for }x < 5\\-x + 10, &\text{for }x>5end{cases}$ $c =$

Explanation:

Step1: Recall limit - existence condition

For $\lim_{x
ightarrow5}f(x)$ to exist, $\lim_{x
ightarrow5^{-}}f(x)=\lim_{x
ightarrow5^{+}}f(x)$.

Step2: Calculate left - hand limit

$\lim_{x
ightarrow5^{-}}f(x)=\lim_{x
ightarrow5^{-}}(\frac{1}{5}x + c)=\frac{1}{5}\times5 + c=1 + c$.

Step3: Calculate right - hand limit

$\lim_{x
ightarrow5^{+}}f(x)=\lim_{x
ightarrow5^{+}}(-x + 10)=-5 + 10 = 5$.

Step4: Set left - hand limit equal to right - hand limit

Set $1 + c=5$.

Step5: Solve for $c$

Subtract 1 from both sides: $c=5 - 1=4$.

Answer:

$4$