Proving using definition complex limit of $(z+1)/(|z|-1)$

Question

What I initially though a simple question has me stumped.

Use the definition of a limit to obtain $$\lim_{z \to 0} \frac{z+1}{|z|-1}$$ Clearly, the limit is $-1$ but I am struggling to prove with the definition. Obviously, I need to demonstrate that $$\left|f(z) + 1 \right| < \epsilon \quad \text{when} \quad 0 < z-z_0<\delta$$ That is $$\left| \frac{z+1}{|z|-1} +1 \right| < \epsilon$$

Now $$\left| \frac{z+1}{|z|-1} +1 \right| = \left| \frac{|z|+z}{|z|-1} \right| \leq \frac{2|z|}{||z|-1|}$$

I just can't seem to put into a useful form. I've also played around with multiplying by the conjugate to get

$$\frac{2|z|||z|+1|}{||z|^2-1|} = \frac{2(|z|^2+|z|)}{||z|^2-1|}$$

But it appears I'm making an even bigger mess of it!

Would really appreciate some pointers! I'm obviously missing something

Thanks

Answer 1

Given $\varepsilon>0$, you have to find $\delta>0$ such that, for $0<|z|<\delta$, $|f(z)+1|<\varepsilon$.

You want to get an upper bound of $1/\bigl||z|-1\bigr|$: if you restrict to $\delta<1/2$, you have, for $|z|<\delta$ (still unspecified), $1-|z|>1/2$, so $0<1/(1-|z|)<2$.

Now, take $\delta=\min(\varepsilon/4,1/2)$; then, if $0<|z|<\delta$ $$ \left|\frac{z+1}{|z|-1}+1\right|= \left|\frac{z+|z|}{|z|-1}\right|\le \frac{2|z|}{1-|z|}<\dots $$