Answer
Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-1|\lt\delta\Rightarrow|f(x)-1|\lt\frac{1}{4}$$
Looking at the graphs, we can see that for values of $f(x)$ to be restricted between $\frac{3}{4}$ and $\frac{5}{4}$ (which means $|f(x)-1|\lt\frac{1}{4}$), $x$ must be placed between $\frac{9}{16}$ and $\frac{25}{16}$.
$|\frac{25}{16}-1|=|\frac{9}{16}|=\frac{9}{16}$
And,
$|1-\frac{9}{16}|=|\frac{7}{16}|=\frac{7}{16}$
Since, $\frac{7}{16} \lt \frac{9}{16},$
$\delta=\frac{7}{16}$
Work Step by Step
Find a $\delta\gt0$ such that for all $x$ $$0\lt |x-1|\lt\delta\Rightarrow|f(x)-1|\lt\frac{1}{4}$$
Looking at the graphs, we can see that for values of $f(x)$ to be restricted between $\frac{3}{4}$ and $\frac{5}{4}$ (which means $|f(x)-1|\lt\frac{1}{4}$), $x$ must be placed between $\frac{9}{16}$ and $\frac{25}{16}$.
And since $|\frac{25}{16}-1|=|\frac{9}{16}|=\frac{9}{16}$, that means $0\lt |x-1|\lt\frac{9}{16}$.
Therefore, $\delta=\frac{9}{16}$ (or any smaller positives number is also fine).
