Question

evaluate the following limit.
\\( \lim_{h \to 0} \frac{\frac{1}{8 + h} - \frac{1}{8}}{h} \\)

rewrite the given limit in a way that can be evaluated by direct substitution.
\\( \lim_{h \to 0} \frac{\frac{1}{8 + h} - \frac{1}{8}}{h} = \lim_{h \to 0} \left( \square \
ight) \\) (simplify your answer)

Explanation:

Step1: Combine the fractions in the numerator

First, we need to combine the two fractions \(\frac{1}{8 + h}\) and \(\frac{1}{8}\) in the numerator. To do this, we find a common denominator, which is \(8(8 + h)\). So we have:
\[
\frac{1}{8 + h}-\frac{1}{8}=\frac{8 - (8 + h)}{8(8 + h)}
\]

Step2: Simplify the numerator

Simplify the numerator of the combined fraction:
\[
\frac{8 - (8 + h)}{8(8 + h)}=\frac{8 - 8 - h}{8(8 + h)}=\frac{-h}{8(8 + h)}
\]

Step3: Divide by \(h\) (multiply by \(\frac{1}{h}\))

Now, we have the original limit as \(\lim_{h
ightarrow0}\frac{\frac{1}{8 + h}-\frac{1}{8}}{h}\), which is the same as \(\lim_{h
ightarrow0}\frac{\frac{-h}{8(8 + h)}}{h}\). Dividing by \(h\) is equivalent to multiplying by \(\frac{1}{h}\), so we get:
\[
\lim_{h
ightarrow0}\frac{-h}{8(8 + h)}\times\frac{1}{h}
\]

Step4: Cancel out \(h\)

We can cancel out the \(h\) terms (assuming \(h
eq0\), which is valid for the limit as \(h
ightarrow0\) since we are approaching \(0\) but not equal to \(0\)):
\[
\lim_{h
ightarrow0}\frac{-1}{8(8 + h)}
\]

Answer:

\(\frac{-1}{8(8 + h)}\) (and then we can directly substitute \(h = 0\) to get \(-\frac{1}{64}\), but for the rewritten form, it's \(\frac{-1}{8(8 + h)}\))