Limits in Complex sequences

Question

Proof that $\displaystyle\lim_{z\to \infty}\dfrac{\log z}{\sqrt[n]{z}}=0$, in Complex numbers.
I have the following question. Can I tell that if the limit $\displaystyle\lim_{|z|\to +\infty}\dfrac{\log |z|}{\sqrt[n]{|z|}}=0$, then our previous limit goes to $0$, as z tends to $\infty$? I think that this is true if our first limit exist. I should proof that with the definition. Any ideas? (Please, don't try to comput the limit with L'Hopital rule because I am trying to understand the definition). Thanks

Answer 1

You may consider the following limit instead. \begin{align*} \lim_{z \to \infty} \left| \frac{\log z}{\sqrt[n]{z}} \right| & = \lim_{z \to \infty}\frac{|\log z|}{|\sqrt[n]{z}|} \\ & = \lim_{z \to \infty}\frac{|\ln |z| + i \operatorname{Arg} (z)|}{| |z|^{1/n} e^{i \operatorname{Arg} (z)/n} |} \\ & = \lim_{z \to \infty}\frac{|\ln |z| + i \operatorname{Arg} (z)|}{|z|^{1/n}} \\ & \leq \lim_{z \to \infty}\frac{|\ln |z|| + |i \operatorname{Arg} (z)|}{|z|^{1/n}} &\text{(By triangle inequality)} \\ & = \lim_{z \to \infty}\frac{\ln |z|}{|z|^{1/n}} + \lim_{z \to \infty}\frac{|\operatorname{Arg} (z)|}{|z|^{1/n}} \end{align*} The first limit can be reduced to the form $$ \lim_{x \to \infty}\frac{\ln x}{x^{1/n}} $$ where $x$ is real. So the limit can be computed to be 0 by using techniques in real analysis. For the second limit, notice that $$ \frac{|\operatorname{Arg} (z)|}{|z|^{1/n}} \leq \frac{\pi}{|z|^{1/n}} \to 0 \text{ as } z \to \infty $$ So the limit is 0 by squeeze theorem. Altogether, we have shown $$ \lim_{z \to \infty} \frac{\log z}{\sqrt[n]{z}} \leq \lim_{z \to \infty} \left| \frac{\log z}{\sqrt[n]{z}} \right| \leq \lim_{z \to \infty}\frac{\ln |z|}{|z|^{1/n}} + \lim_{z \to \infty}\frac{|\operatorname{Arg} (z)|}{|z|^{1/n}} = 0 + 0 = 0. $$ Therefore the original limit is also 0 by squeeze theorem.