Question

find $lim_{h
ightarrow0}\frac{f(4 + h)-f(4)}{h}$ if $f(x)=x^{2}+2$.
$lim_{h
ightarrow0}\frac{f(4 + h)-f(4)}{h}=square$ (simplify your answer.)

Explanation:

Step1: Find f(4 + h) and f(4)

First, find f(4 + h) by substituting x = 4+h into f(x)=x² + 2. So f(4 + h)=(4 + h)²+2=16 + 8h+h²+2=h² + 8h+18. And f(4)=4²+2=16 + 2=18.

Step2: Substitute into the limit expression

Substitute f(4 + h) and f(4) into $\lim_{h
ightarrow0}\frac{f(4 + h)-f(4)}{h}$. We get $\lim_{h
ightarrow0}\frac{(h^{2}+8h + 18)-18}{h}=\lim_{h
ightarrow0}\frac{h^{2}+8h}{h}$.

Step3: Simplify the expression

Factor out h from the numerator: $\lim_{h
ightarrow0}\frac{h(h + 8)}{h}$. Cancel out the common - factor h (since h→0 but h≠0), we get $\lim_{h
ightarrow0}(h + 8)$.

Step4: Evaluate the limit

As h approaches 0, substitute h = 0 into h + 8. So $\lim_{h
ightarrow0}(h + 8)=8$.

Answer:

8