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
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