Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 3

Determine a number $\delta$ such that if $| x - 4| < \delta$ then $|\sqrt{x}-2| < 0.4$ by using the graph of $f(x) = \sqrt{x}$








First, we will get the values of $x$ that intersect at the given curve to their corresponding $y$ values. Let $x_L$ and $x_R$
are the values of $x$ from the left and right of 4 respectively.

$
\begin{equation}
\begin{aligned}
y & = \sqrt{x_L} &
y & = \sqrt{x_R}\\

1.6 & = \sqrt{x_L} &
2.4 & = \sqrt{x_R}\\

(1.6)^2 & = (\sqrt{x_L})^2 &
(2.4)^2 & = (\sqrt{x_R})^2\\

x_L & = (1.6)^2 = 2.56&
x_R & = (2.4)^2 = 5.76\\

x_L & = 2.56 &
x_R & = 5.76
\end{aligned}
\end{equation}
$


Now, we can determine the value of $\delta$ by checking the values of $x$ that would give a smaller distance to 4.


$
\begin{equation}
\begin{aligned}
4 - x_L & = 4 - 2.56 = 1.44\\
x_R - 4 & = 5.76 - 4 = 1.76
\end{aligned}
\end{equation}
$


Hence,
$\quad \delta \leq 1.44$

This means that by keeping $x$ within $1.44$ of $4$, we are able to keep $f(x)$ within $0.4$ of $2$.

Although we chose $\delta = 1.44$, any smaller positive value of $\delta$ would also have work.