Question

$\lim_{h \to 0} \frac{\frac{2}{5+h} - \frac{2}{5}}{h}$
answer: $\square$

Explanation:

Step1: Simplify the numerator

First, we simplify the numerator \(\frac{2}{5 + h}-\frac{2}{5}\). Find a common denominator, which is \(5(5 + h)\). Then we have:
\[

$$\begin{align*} \frac{2}{5 + h}-\frac{2}{5}&=\frac{2\times5-2\times(5 + h)}{5(5 + h)}\\ &=\frac{10-(10 + 2h)}{5(5 + h)}\\ &=\frac{10 - 10-2h}{5(5 + h)}\\ &=\frac{-2h}{5(5 + h)} \end{align*}$$

\]

Step2: Substitute back into the limit

Now our limit becomes:
\[
\lim_{h
ightarrow0}\frac{\frac{-2h}{5(5 + h)}}{h}
\]
We can rewrite the complex fraction as \(\lim_{h
ightarrow0}\frac{-2h}{5(5 + h)}\times\frac{1}{h}\). The \(h\) terms (where \(h
eq0\)) will cancel out, leaving us with:
\[
\lim_{h
ightarrow0}\frac{-2}{5(5 + h)}
\]

Step3: Evaluate the limit

Now we can substitute \(h = 0\) into the expression \(\frac{-2}{5(5 + h)}\) since the function is now continuous at \(h = 0\) (after canceling the \(h\) terms). So we get:
\[
\frac{-2}{5(5+0)}=\frac{-2}{25}
\]

Answer:

\(-\frac{2}{25}\)