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 a curious limit JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 04/14/2011, 08:01 PM (This post was last modified: 04/14/2011, 08:09 PM by JmsNxn.) 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 $v =\pi (1 + 2k) \,\,\,\,\{k \,\epsilon\, N\}$ nuninho1980 Fellow Posts: 95 Threads: 6 Joined: Apr 2009 04/14/2011, 08:10 PM (This post was last modified: 04/14/2011, 08:11 PM by nuninho1980.) attention: h -> o is bad but yes h -> 0. because 'o' isn't number and is letter. lol JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 04/14/2011, 08:21 PM (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 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/14/2011, 10:14 PM (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. JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 04/14/2011, 10:55 PM (This post was last modified: 04/14/2011, 10:56 PM by JmsNxn.) 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? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/15/2011, 07:14 AM (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. JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 04/15/2011, 05:09 PM (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. JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 04/16/2011, 07:22 PM is there any way of letting h approach zero such that: $\lim_{h\to\0} (1-e^{vi})ln(h) = 0$? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/16/2011, 07:41 PM (This post was last modified: 04/16/2011, 08:18 PM by bo198214.) (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 $1=e^{vi}$). JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 04/16/2011, 07:48 PM (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 « Next Oldest | Next Newest »

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