Question

find the following limit or state that it does not exist. (limlimits_{h \to 0} \frac{\frac{8}{3+h} - \frac{8}{3}}{h}) simplify the given limit (limlimits_{h \to 0} \frac{\frac{8}{3+h} - \frac{8}{3}}{h} = limlimits_{h \to 0} left( \frac{-8}{3(3+h)}
ight)) (simplify your answer.) evaluate the limit, if possible. select the correct choice below and, if necessary, fill in the answer box to complete your choice. (\bigcirc) a. (limlimits_{h \to 0} \frac{\frac{8}{3+h} - \frac{8}{3}}{h} = square) (type an exact answer.) (\bigcirc) b. the limit does not exist.

Explanation:

Step1: Simplify the numerator

First, we simplify the numerator \(\frac{8}{3 + h}-\frac{8}{3}\). Find a common denominator, which is \(3(3 + h)\). Then we have:
\[

$$\begin{align*} \frac{8}{3 + h}-\frac{8}{3}&=\frac{8\times3-8\times(3 + h)}{3(3 + h)}\\ &=\frac{24-24 - 8h}{3(3 + h)}\\ &=\frac{-8h}{3(3 + h)} \end{align*}$$

\]
So the original limit becomes \(\lim_{h
ightarrow0}\frac{\frac{-8h}{3(3 + h)}}{h}\).

Step2: Cancel out \(h\)

We can cancel out the \(h\) in the numerator and the denominator (since \(h
eq0\) when taking the limit as \(h
ightarrow0\)):
\[
\lim_{h
ightarrow0}\frac{\frac{-8h}{3(3 + h)}}{h}=\lim_{h
ightarrow0}\frac{-8}{3(3 + h)}
\]

Step3: Substitute \(h = 0\)

Now we substitute \(h = 0\) into the expression \(\frac{-8}{3(3 + h)}\):
\[
\frac{-8}{3(3+0)}=\frac{-8}{9}
\]

Answer:

A. \(\lim\limits_{h
ightarrow0}\frac{\frac{8}{3 + h}-\frac{8}{3}}{h}=-\frac{8}{9}\)