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:

$$
LATEXBLOCK0
$$

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:

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