text: # Tetration Forum
  
  Full Version: a curious limit
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  I'm wondering if the following limit is non-zero; v E R
  
  $\lim_{h\to\0}h^{1-e^{vi}}$
  
  and if so, what is it equal to? Thanks
  
  I know it doesn't converge for $v =\pi (1 + 2k) \,\,\,\,\{k \,\epsilon\, N\}$
  attention: h -> o is bad but yes h -> 0. because 'o' isn't number and is letter. lol
  (04/14/2011, 08:10 PM)nuninho1980 Wrote: [ -> ]attention: h -> o is bad but yes h -> 0. because 'o' isn't number and is letter. lol
  
  no no, I put 0, but latex just designs it to look like o
  (04/14/2011, 08:01 PM)JmsNxn Wrote: [ -> ]I'm wondering if the following limit is non-zero; v E R
  
  $\lim_{h\to\0}h^{1-e^{vi}}$
  
  and if so, what is it equal to? Thanks
  
  I know it doesn't converge for $v =\pi (1 + 2k) \,\,\,\,\{k \,\epsilon\, N\}$
  
  The powers with non-integer exponents are not uniquely defined in the complex plane.
  In your case you would need to put:
  
  $\lim_{h\to\0} e^{(1-e^{vi})\log(h)}$
  
  But then the standard logarithm has a cut on $(-\infty,0]$, which is quite arbitrary: one could put a cut however one likes. For example $h$ could spiral around 0, while moving towards 0 and would increase/decrease its imaginary part by $2\pi i$ in each round.
  I guess it really depends on how $h$ approaches 0.
  let's take the limit from positive (keep it simple first)
  
  so:
  $\lim_{h\to\0^+}\, (1-e^{vi})ln(h)\, =\, f(v)$
  
  Is there any way of re-expressing this limit?
  (04/14/2011, 10:55 PM)JmsNxn Wrote: [ -> ]let's take the limit from positive (keep it simple first)
  
  so:
  $\lim_{h\to\0^+}\, (1-e^{vi})ln(h)\, =\, f(v)$
  
  Is there any way of re-expressing this limit?
  
  But then its not difficult, since $\ln(h)\to -\infty$ on the reals, the whole limit goes to (complex) $\infty$ except for $1-e^{vi}=0$, for which the whole limit is 0.
  (04/15/2011, 07:14 AM)bo198214 Wrote: [ -> ]But then its not difficult, since $\ln(h)\to -\infty$ on the reals, the whole limit goes to (complex) $\infty$ except for $1-e^{vi}=0$, for which the whole limit is 0.
  
  Alright, how about
  
  $\lim_{h\to\0^{v+}} (1-e^{vi})ln(h)$
  
  where $\lim_{h\to\0^{v+}}$ is taken to mean approaching along the $e^{vi}$ axis.
  
  I think it's the equivalent of:
  $= \lim_{h\to\0^{+}} (1-e^{vi})ln(he^{vi})$
  $= \lim_{h\to\0^{+}} (1-e^{vi})(ln(h) + vi)$
  which I guess converges to negative infinity again, except for 1-e^{vi}=0
  
  hmm, seems this is less interesting than I thought.
  is there any way of letting h approach zero such that:
  
  $\lim_{h\to\0} (1-e^{vi})ln(h) = 0$?
  (04/16/2011, 07:22 PM)JmsNxn Wrote: [ -> ]is there any way of letting h approach zero such that:
  
  $\lim_{h\to\0} (1-e^{vi})ln(h) = 0$?
  
  The logarithm of $z=r e^{i\phi}$ is $\log( r)+i\phi$.
  So regardless how you approach 0, i.e. $r\to 0$, you will allways have that $|\log(z)|=\sqrt{\log( r)^2+\phi^2}\to \infty$.
  So the answer is no (except $1=e^{vi}$).
  
  (04/16/2011, 07:41 PM)bo198214 Wrote: [ -> ]The logarithm of $z=r e^{i\phi}$ is $\log( r)+i\phi$.
  So regardless how you approach 0, i.e. $r\to 0$, you will allways have that $|\log(z)|=\sqrt{\log( r)^2+\phi^2}\to \infty$.
  So the answer is no (except $1=e^{vi}$).
  
  that's what I thought
  Pages: 1 2
explanation: This document is a discussion thread from a math forum, specifically the Tetration Forum. The main content revolves around a user, JmsNxn, asking a question about a limit involving complex numbers and exponential functions. The question is:
  
  \[
  \lim_{h \to 0} h^{1 - e^{vi}}
  \]
  
  The user is curious if this limit is non-zero and, if so, what its value is. The discussion includes responses from other users, such as nuninho1980 and bo198214, who provide insights and corrections to the notation and the approach to solving the problem. The thread also explores different ways of approaching the limit and the conditions under which the limit converges or diverges.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Tetration Forum
  
  Full Version: a curious limit
  You're currently viewing a stripped down version of our content. View the full version with proper formatting.
  Pages: 1 2
  I'm wondering if the following limit is non-zero; v E R
  
  $\lim_{h\to\0}h^{1-e^{vi}}$
  
  and if so, what is it equal to? Thanks
  
  I know it doesn't converge for $v =\pi (1 + 2k) \,\,\,\,\{k \,\epsilon\, N\}$
  attention: h -> o is bad but yes h -> 0. because 'o' isn't number and is letter. lol
  (04/14/2011, 08:10 PM)nuninho1980 Wrote: [ -> ]attention: h -> o is bad but yes h -> 0. because 'o' isn't number and is letter. lol
  
  no no, I put 0, but latex just designs it to look like o
  (04/14/2011, 08:01 PM)JmsNxn Wrote: [ -> ]I'm wondering if the following limit is non-zero; v E R
  
  $\lim_{h\to\0}h^{1-e^{vi}}$
  
  and if so, what is it equal to? Thanks
  
  I know it doesn't converge for $v =\pi (1 + 2k) \,\,\,\,\{k \,\epsilon\, N\}$
  
  The powers with non-integer exponents are not uniquely defined in the complex plane.
  In your case you would need to put:
  
  $\lim_{h\to\0} e^{(1-e^{vi})\log(h)}$
  
  But then the standard logarithm has a cut on $(-\infty,0]$, which is quite arbitrary: one could put a cut however one likes. For example $h$ could spiral around 0, while moving towards 0 and would increase/decrease its imaginary part by $2\pi i$ in each round.
  I guess it really depends on how $h$ approaches 0.
  let's take the limit from positive (keep it simple first)
  
  so:
  $\lim_{h\to\0^+}\, (1-e^{vi})ln(h)\, =\, f(v)$
  
  Is there any way of re-expressing this limit?
  (04/14/2011, 10:55 PM)JmsNxn Wrote: [ -> ]let's take the limit from positive (keep it simple first)
  
  so:
  $\lim_{h\to\0^+}\, (1-e^{vi})ln(h)\, =\, f(v)$
  
  Is there any way of re-expressing this limit?
  
  But then its not difficult, since $\ln(h)\to -\infty$ on the reals, the whole limit goes to (complex) $\infty$ except for $1-e^{vi}=0$, for which the whole limit is 0.
  (04/15/2011, 07:14 AM)bo198214 Wrote: [ -> ]But then its not difficult, since $\ln(h)\to -\infty$ on the reals, the whole limit goes to (complex) $\infty$ except for $1-e^{vi}=0$, for which the whole limit is 0.
  
  Alright, how about
  
  $\lim_{h\to\0^{v+}} (1-e^{vi})ln(h)$
  
  where $\lim_{h\to\0^{v+}}$ is taken to mean approaching along the $e^{vi}$ axis.
  
  I think it's the equivalent of:
  $= \lim_{h\to\0^{+}} (1-e^{vi})ln(he^{vi})$
  $= \lim_{h\to\0^{+}} (1-e^{vi})(ln(h) + vi)$
  which I guess converges to negative infinity again, except for 1-e^{vi}=0
  
  hmm, seems this is less interesting than I thought.
  is there any way of letting h approach zero such that:
  
  $\lim_{h\to\0} (1-e^{vi})ln(h) = 0$?
  (04/16/2011, 07:22 PM)JmsNxn Wrote: [ -> ]is there any way of letting h approach zero such that:
  
  $\lim_{h\to\0} (1-e^{vi})ln(h) = 0$?
  
  The logarithm of $z=r e^{i\phi}$ is $\log( r)+i\phi$.
  So regardless how you approach 0, i.e. $r\to 0$, you will allways have that $|\log(z)|=\sqrt{\log( r)^2+\phi^2}\to \infty$.
  So the answer is no (except $1=e^{vi}$).
  
  (04/16/2011, 07:41 PM)bo198214 Wrote: [ -> ]The logarithm of $z=r e^{i\phi}$ is $\log( r)+i\phi$.
  So regardless how you approach 0, i.e. $r\to 0$, you will allways have that $|\log(z)|=\sqrt{\log( r)^2+\phi^2}\to \infty$.
  So the answer is no (except $1=e^{vi}$).
  
  that's what I thought
  Pages: 1 2
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: This document is a discussion thread from a math forum, specifically the Tetration Forum. The main content revolves around a user, JmsNxn, asking a question about a limit involving complex numbers and exponential functions. The question is:
  
  \[
  \lim_{h \to 0} h^{1 - e^{vi}}
  \]
  
  The user is curious if this limit is non-zero and, if so, what its value is. The discussion includes responses from other users, such as nuninho1980 and bo198214, who provide insights and corrections to the notation and the approach to solving the problem. The thread also explores different ways of approaching the limit and the conditions under which the limit converges or diverges.
  
  Label: question-and-answer