Delta Epsilon Proof $1/z$ Use the definition to prove that:
$$\lim_{z\to 1+i} \frac{1}{z} = \frac{1}{1+i}$$
So far this is what I have.
$$\left|\frac{1}{z} - \frac{1}{1+i}\right| < \epsilon$$
Since 
$$\left|\frac{1}{z} - \frac{1}{1+i}\right| = \left|\frac{z-(1+i)}{z(1+i)}\right|$$
$$= \frac{|z-(1+i)|}{|z|\sqrt{2}}$$
$$\frac{|z-(1+i)|}{|z|\sqrt{2}}< \epsilon$$
I don't know where to go from here. I'm not sure what to do to get rid of the $z$ in the denominator, looking at other questions it looks like I am expected to restrict it to something but I don't really understand. 
 A: Since you're trying to get close to $1+i$ which lies outside the unit circle, maybe make the assumption that $|z|>1$. This way you can use $\frac{1}{|z|}<1$.
Now that you have this, just pick the right  $\delta$ such that $|z-1-i|<\delta$
A: Ravi's answer is right, but perhaps adding a few things to it would help.
The problem is to make $\delta$ small enough for a given $\varepsilon$.  If some particular value of $\delta$ is small enough, then so is any smaller $\delta$ (as long as it is positive).  Thus it is harmless to decide in advance always to choose $\delta$ to be $\le1/10$.  And as we will see, it can be useful.
If the distance between $1+i$ and $z$ is less than $1/10$, then $|z|>1$. (Check that.)
If $|z|>1$ then $\dfrac{|z-(1+i)|}{|z|\sqrt{2}} < \dfrac{|z-(1+i)|}{\sqrt{2}}.$
And $\dfrac{|z-(1+i)|}{\sqrt{2}} < \varepsilon$ if $|z-(1+i)| < \varepsilon\cdot\sqrt 2.$
Thus if one lets $$\delta = \min\left\{ \dfrac 1 {10},\  \sqrt 2 \cdot \varepsilon \right\}, \tag 1$$ that will serve.
The proof should not be written in the way in which you wrote in your posted question. What you wrote is scratchwork needed to figure out how to write the proof.  You should show how $(1)$ above leads to the conclusion that $\dfrac{|z-(1+i)|}{|z|\sqrt{2}} < \varepsilon.$
