Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 4

Find a number $\delta$ such that if $|x -1| < \delta $ then $\displaystyle |x^2 - 1| < \frac{1}{2}$ using the given graph of $f(x) = x^2$









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 1 respectively.

$
\begin{equation}
\begin{aligned}
y & = (x_L)^2 &
y & = (x_R)^2\\

0.5 & = (x_L)^2 &
1.5 & = (x_R)^2\\

\sqrt{0.5} & = \sqrt{(x_L})^2 &
\sqrt{1.5} & = \sqrt{(x_R})^2\\

x_L & = \sqrt{0.5} &
x_R & = \sqrt{1.5} \\

x_L & = 0.7071 &
x_R & = 1.2247
\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 1.


$
\begin{equation}
\begin{aligned}
1 - x_L & = 1 - 0.7071 = 0.2929\\
x_R - 1 & = 1.2247 - 1 = 0.2247
\end{aligned}
\end{equation}
$


Hence,
$\quad \delta \leq 0.2247$

This means that by keeping $x$ within $0.2247$ of $1$, we are able to keep $f(x)$ within $0.5$ of $1$.

Although we chose $\delta = 0.2247$, any smaller positive value of $\delta$ would also have work.