Question

evaluate
$lim_{h \to 0} \frac{f(3 + h) - f(3)}{h}$,
where $f(x) = 5x + 1$. if the limit does not exist enter dne.
limit = 18

Explanation:

Step1: Compute $f(3+h)$

Substitute $x=3+h$ into $f(x)$.
$f(3+h)=5(3+h)+1=15+5h+1=16+5h$

Step2: Compute $f(3)$

Substitute $x=3$ into $f(x)$.
$f(3)=5(3)+1=15+1=16$

Step3: Substitute into limit expression

Replace $f(3+h)$ and $f(3)$ in the limit.
$\lim_{h \to 0} \frac{(16+5h)-16}{h} = \lim_{h \to 0} \frac{5h}{h}$

Step4: Simplify and evaluate limit

Cancel $h$ (valid for $h
eq0$), then take limit.
$\lim_{h \to 0} 5 = 5$

Answer:

5