Answer
$\delta=0.4$
Work Step by Step
1) From the definition of limit, the statement $$\lim_{x\to c}f(x)=L$$
means for each $\varepsilon\gt0$ there exists a $\delta\gt0$ such that if
$$0\lt|x-c|\lt\delta$$ then
$$|f(x)-L|\lt\varepsilon$$
2) The question asks to find $\delta$ such that if $0\lt|x-2|\lt\delta$ then $|f(x)-3|\lt0.4$, with $f(x)=x+1$
So we have already been given $\varepsilon=0.4$. To find $\delta$, we need to find a connection between $|x-2|$ and $|f(x)-3|$.
To do so, we replace $f(x)=x+1$ into $|f(x)-3|$.
$$|f(x)-3|=|(x+1)-3|=|x-2|$$
Since $|f(x)-3|\lt0.4$, now $|x-2|\lt0.4$
Therefore as we need to find $\delta$ for $0\lt|x-2|\lt\delta$, we can pick $\delta=0.4$.
3) Try $\delta=0.4$ back again
For $0\lt|x-2|\lt0.4$, we have $$|f(x)-3|=|(x+1)-3|=|x-2|\lt0.4$$
So $\delta=0.4$ is a right choice.
*From the graph, we can also see that for x-values within $0.4$ of $2$ $(x\ne2)$, y-values are within $0.4$ of $3$.
