Answer
$$0.1,\ \ 0.0990,\ \ \ 0.00098$$
Work Step by Step
Given $$\sqrt{26}-\sqrt{25}$$
Consider $f(x) =\sqrt{x} $ and $ a =25$, $\Delta x= 1 $, since
\begin{align*}
f'(x)&= \frac{1}{2\sqrt{x}} \\
f'(25)&= 0.1
\end{align*}
Then the linear approximate is given by
\begin{align*}
\Delta f& \approx . f^{\prime}(a) \Delta x\\
&= (0.1)( 1)=0.1
\end{align*}
and the actual change is given by
\begin{align*}
\Delta f&=f(a+\Delta x)-f(a)\\
&=f(26)-f(25)\\
& \approx 0.0990
\end{align*}
Hence the error is given by
$$|0.0990-0.1|=0.000980 $$
