Question

1. $limlimits_{h \to 0} \frac{(5 + h)^2 - 25}{h}$

Explanation:

Step1: Expand the numerator

First, we expand \((5 + h)^2\) using the formula \((a + b)^2 = a^2 + 2ab + b^2\). Here, \(a = 5\) and \(b = h\), so \((5 + h)^2 = 25 + 10h + h^2\). Then the numerator becomes \((25 + 10h + h^2) - 25\).
\[

$$\begin{align*} (5 + h)^2 - 25&=25 + 10h + h^2 - 25\\ &=10h + h^2 \end{align*}$$

\]

Step2: Simplify the fraction

Now we substitute the expanded numerator back into the original limit expression. So we have \(\lim_{h \to 0} \frac{10h + h^2}{h}\). We can factor out an \(h\) from the numerator: \(10h + h^2 = h(10 + h)\). Then the fraction becomes \(\frac{h(10 + h)}{h}\). Since \(h \to 0\) but \(h
eq 0\) (we are taking the limit as \(h\) approaches 0, not evaluating at \(h = 0\)), we can cancel out the \(h\) terms.
\[
\lim_{h \to 0} \frac{h(10 + h)}{h}=\lim_{h \to 0} (10 + h)
\]

Step3: Evaluate the limit

Now we evaluate the limit as \(h\) approaches 0. We substitute \(h = 0\) into the expression \(10 + h\).
\[
\lim_{h \to 0} (10 + h)=10 + 0 = 10
\]

Answer:

\(10\)