$\displaystyle \displaystyle\lim_{z\to i}\frac{z^2-1}{z^2+1}=\frac{(z+1)(z-1)}{(z+i)(z-i)}$
This limit is giving me a little trouble.
I tried factoring but I wasn't able to go anywhere from there.
Anyone have any suggestions?
Thanks.
$\displaystyle \displaystyle\lim_{z\to i}\frac{z^2-1}{z^2+1}=\frac{(z+1)(z-1)}{(z+i)(z-i)}$
This limit is giving me a little trouble.
I tried factoring but I wasn't able to go anywhere from there.
Anyone have any suggestions?
Thanks.
Suppose it were $\displaystyle \displaytype \lim_{x\to 0}\frac{1}{x}$. How would you show that goes to infinity?
For z close to i, $\displaystyle z^2- 1$ will be close to -2 and $\displaystyle z+ i$ will be close to 2i so
$\displaystyle \frac{z^2- 1}{(z+i)(z-i)}$ will be close to $\displaystyle \frac{i}{2i(z- i)}= \frac{1}{2}\frac{1}{z- i}$. You want to show that, given any M, there exist r such that if $\displaystyle |z- i|< r$, then $\displaystyle \frac{1}{2}\left|\frac{1}{z-i}\right|< M$. That should be pretty straight forward.