Answer
$\delta = \frac{1}{3}$
Work Step by Step
First we need to find the function values around the limit where $\left|\frac{2x}{x^2+4}-0.4\right| < 0.1$
These work out to:
$0.4 - 0.1 = 0.3$
$0.4 + 0.1 = 0.5$
Now using the graph we can find $x$-values corresponding to these $y$-values.
for $f(x) = 0.3, x = \frac{2}{3}$
for $f(x) = 0.5, x = 2$
Finally, plugging these into our inequality we get:
$ 0 < \left|x-1 \right| < \delta$
$ 0 < \left|\frac{2}{3}-1 \right| < \delta$
$ 0 < \frac{1}{3}< \delta$
and
$ 0 < \left|x-1 \right| < \delta$
$ 0 < \left|2-1 \right| < \delta$
$ 0 <1< \delta$
Choose the smaller value for $\delta$ and we are left with
$ \delta = \frac{1}{3}$
