You want to prove that $\lim\limits_{x\to 1}(x+4) = 5$ using $\epsilon$-$\delta$.
Let $\epsilon\gt 0$. We need to prove that there exists a $\delta\gt 0$ such that
If $0\lt |x-1|\lt \delta$ then $|f(x)-5|\lt \epsilon$.
Now, we want to think a bit: how will the size of $|x-1|$ affect the size of $|f(x)-5|$? Since $f(x)=x+4$, we notice that $|f(x)-5| = |(x+4)-5| = |x-1|$; that is, the size of $|f(x)-5|$ is equal to the size of $|x-1|$. So in order to make sure that $|f(x)-5|\lt \epsilon$, it is enough to require that $|x-1|\lt\epsilon$.
Thus, we can select $\delta=\epsilon$. Then $\delta\gt 0$, and if $0\lt |x-1|\lt\delta$, then it will follow that $|f(x)-5|\lt\epsilon$.
Thus, for all $\epsilon\gt 0$ there exists a $\delta\gt 0$ (namely, $\delta=\epsilon$) with the property that if $0\lt |x-1|\lt \delta$, then $|f(x)-5|\lt \epsilon$. This proves that $\lim\limits_{x\to 1}f(x) = 5$, as desired. $\Box$
That's what you have, only with lots of words thrown in in-between...
Remember the definition of limit,
$$\forall \epsilon\; \exists \delta : |x-a|<\delta \rightarrow |f(x) - L|<\epsilon $$
Now if $L$ is not the limit negate the definition, that is
$$ \exists \epsilon\; \forall \delta: \;\; \exists x \text{ s.t. } |x-a| <\delta\text{ and } |f(x) - L| \geq \epsilon $$
In your case, as some people already mentioned take $\epsilon =1/2$, now for ever $\delta$ we need to find an $x$ such that $|x-4|<\delta$ and $|9-x-4|\geq 1/2$. Finding such $x$ isn't that difficult, just take $0<r<\min\{\delta, 1/2\} $ and consider $x=4+r$. To check we substitute in the inequalities:
$$|x-4|=|r|<\delta$$
And
$$|9-x-4|=|1-r|>1/2= \epsilon $$
(sorry for the awful format, I'm on my phone)
Best Answer
Factoring $i$ out of $1+zi$ gives $i(z+\dfrac{1}{i})=i(z-i)$.