if lim g(z)=c $\ne 0$ and lim f(z) = $\infty$ then lim g(x)f(x) =$\infty$ (Proof)

Question

Note z and c are complex.

$\forall \epsilon > 0$ $\exists M>0$ s.t $\forall |z|\geq M$: $|g(z)-c| < \frac{|c|}{2}$ i.e. take $\epsilon =\frac{|c|}{2}$

This implies that : $||g(z)|-|c|| <\frac{|c|}{2}$ Therefore: $ |c|-\frac{|c|}{2}\leq|g(z)|$ $\forall |z|\geq M$

$\forall l > 0$ $\exists N>0$ s.t $\forall |z|\geq N$ $|f(z)| >\frac{l}{ |c|-\frac{|c|}{2}}$

Therefore:

$\forall l > 0$ $\exists M_0 = max\{M,N\}>0$ s.t $\forall |z|\geq M_0$: $|f(z)g(z)| >\frac{l}{ |c|-\frac{|c|}{2}}$$(|c|-\frac{|c|}{2})>l$

Is this right?

Answer 1

It's not specified whether the limit is for $z\to\infty$ or for $z\to z_0$, but your logic is good for the former case.

The idea is that $g$ is bounded below in a neighborhood of the point the limit is taken at, so $|g|$ is bounded as well.

Let's do it for $z\to\infty$, you can do similarly for $z\to z_0$.

There exists $M>0$ such that, for $|z|>M$, $|g(z)-c|<\frac{|c|}{2}$, so also $$ \bigl||g(z)|-|c|\bigr|<\frac{|c|}{2} $$ and so $-\frac{|c|}{2}<|g(z)|-|c|<\frac{|c|}{2}$, which implies $|g(z)|>|c|/2$.

Now let $K>0$. There exists $N>0$ so that, for $|z|>N$, $$ |f(z)|>\frac{2K}{|c|} $$ and it's not restrictive to assume $N>M$. Hence, for $|z|>N$, $$ |f(z)g(z)|\ge|f(z)|\frac{|c|}{2}>\frac{2K}{|c|}\frac{|c|}{2}=K $$

For $z\to z_0\in\mathbb{C}$ it's very similar.

Answer 2

Looks okay, but a bit messy.

From $|g(z)-c|<\frac12|c|$ it follows directly that $|g(z)|\geq\frac12|c|$.

This because $|g(z)|<\frac12|c|$ combined with $|g(z)-c|<\frac12|c|$ leads to: $$|c|=|c-g(z)+g(z)|\leq|c-g(z)|+|g(z)|<\frac12|c|+\frac12|c|=|c|$$which is absurd.