How to compute the limit of a complex function below? Thanks.
For $b>a>0$, $x\in(-\infty,-a)$, $r>0$, $s\in\mathbb{R}$, and $i=\sqrt{-1}\,$, let
\begin{equation*}
f_{a,b;s}(x+ir)=
\begin{cases}
\ln\dfrac{(x+ir+b)^s-(x+ir+a)^s}s, & s\ne0;\\
\ln\ln\dfrac{x+ir+b}{x+ir+a}, & s=0.
\end{cases}
\end{equation*}
Compute the limit
\begin{equation*}
\lim_{r\to0^+}f_{a,b;s}(x+ir).
\end{equation*}
Before computing the limit one has to fix the precise meaning of the expression. The functions $g(z) = z^s$ and $h(z) = \ln(z)$ are multi-valued, so the question will be correctly posed only after selecting appropriate branches of all expressions that include these functions. Outside of this I see no difficulties.
Could you please check your question if you mean
\lim_{s\to0^+}f_{a,b;s}(x+ir).
instead of
\lim_{r\to0^+}f_{a,b;s}(x+ir).?
For me, the variable is s, not r if you look at the definition of the function. Then the limit is
\ln\ln\dfrac{x+ir+b}{x+ir+a
and so the function defined there is continuous for s at s=0+.
Cite
1 Recommendation
Alip Mohammed
McGill University
I think, L'Hopitals rule is applicable for the function you have keeping everything as parameter, considering s as variable if I am not wrong.
Cite
Sjoerd W. Rienstra
Eindhoven University of Technology
Write your non-integer powers with logs: z^s = exp(s log(z)) and define all logs the way you want or need. If you use principal value logs, then you seem to have no problem if x+b0 and s->0 in any order. However, if x+b>0, the limit s->0 after the limit r->0 does not exist.
Cite
1 Recommendation
Feng Qi
Texas
My question is to compute $\lim_{r\to0^+}f_{a,b;s}(x+ir)$.
- Badges
- Science topic
- Similar topics
- Mathematics