/sci/ - Science & Math » Thread #12807190

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Explain mathfags

Anonymous Mon 08 Mar 17:33:31 2021 No.12807190 View Reply Original Report

Quoted By: >>12807239 >>12807259 >>12807305 >>12807329 >>12807339 >>12807394 >>12807601

$\lim_{x\to\infty} x^0$ is indefinite but for any number N, $N^0 = 1$ , why shouldn't big values of X act the same way?

Anonymous

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Anonymous Mon 08 Mar 2021 17:42:16 No.12807239 Report

Quoted By: >>12807305

>>12807190
$0^0$ stfu

Anonymous

Anonymous Mon 08 Mar 2021 17:44:05 No.12807247 Report

Quoted By:

Infinity is not a number. Faggot.

Anonymous

Anonymous Mon 08 Mar 2021 17:47:17 No.12807259 Report

Quoted By: >>12807293

>>12807190
>limx?x0 is indefinite
I don't think it is

Anonymous

Anonymous Mon 08 Mar 2021 17:51:17 No.12807279 Report

Quoted By:

$0^0^0^0^0^0^0^0^0^0^0^0^0^0^0^0^0^0^0^0^0^0^0^0^0^0^0^0^0^0^$

Anonymous

Anonymous Mon 08 Mar 2021 17:54:46 No.12807293 Report

Quoted By: >>12807308 >>12807388

>>12807259
$\lim_{x\to\infty} \left(1 + \frac{1}{x}\right)^x \neq \lim_{x\to\infty} \left(1 + \frac{2}{x}\right)^{2x}$

Anonymous

Anonymous Mon 08 Mar 2021 17:56:40 No.12807305 Report

Quoted By:

>>12807190
>>12807239
Thanks doc

Anonymous

Anonymous Mon 08 Mar 2021 17:57:03 No.12807308 Report

Quoted By: >>12807323

>>12807293
x=0=2x
try again

Anonymous

Anonymous Mon 08 Mar 2021 18:02:14 No.12807323 Report

Quoted By: >>12807398

>>12807308
nigga open your eyes when replying

Anonymous

Anonymous Mon 08 Mar 2021 18:03:03 No.12807329 Report

Quoted By:

>>12807190
>lim x? x^0 is indefinite
That's not even meaningful. lim x? x^0 = 1. Did you mean to say that ?^0 is an indefinite form, perhaps?

Anonymous

Anonymous Mon 08 Mar 2021 18:06:25 No.12807339 Report

Quoted By:

>>12807190
the limit is 1, wtf

Anonymous

Anonymous Mon 08 Mar 2021 18:15:40 No.12807388 Report

Quoted By:

>>12807293
This has nothing to do at all with the claim in OP. Are you sure you're in the right thread? If you are, consider studying what an indeterminate form is before making threads about it.

Anonymous

Anonymous Mon 08 Mar 2021 18:17:16 No.12807394 Report

Quoted By:

>>12807190
my gf is so hot

Anonymous

Anonymous Mon 08 Mar 2021 18:17:58 No.12807398 Report

Quoted By:

>>12807323
what does your counterexample have to do with the zeroth power of x or f(x) as x->??

it's two f(x) that both -> 1 as x->?, but have different limits when exponentiated by two other f(x) that ->? as x->?. how does that relate?

Anonymous

Anonymous Mon 08 Mar 2021 18:26:57 No.12807441 Report

Quoted By: >>12808243

Let $a$ be any non-zero complex number. If $\rm Log$ denotes the principal value, we have:
$0^0\;=\;\lim_{z\to0}z^{z\,\mathrm{Log}\,a}\;=\;\lim_{z\to0}\mathrm e^{z\,\mathrm{Log}\,a\,\mathrm{Log}\,z}\;=\;\mathrm e^{\mathrm{Log}\,a\,\lim_{z\to0}z\,\mathrm{Log}\,z}\;=\;\mathrm e^{\mathrm{Log}\,a}\;=\;a.$
We have shown that $\forall a\,\in\,\mathbf C^*,\,0^0\;=\;a$ . In other words, OP is a faggot.

Anonymous

Anonymous Mon 08 Mar 2021 18:57:50 No.12807601 Report

Quoted By:

>>12807190
Learn how limits work retard. The limit is well defined and 1 precisely because of what you said.

Anonymous

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Anonymous Mon 08 Mar 2021 21:51:08 No.12808243 Report

Quoted By:

>>12807441
You have assumed, in your penultimate step, that
$\lim_{z\to 0} z \log z = 1$
This is inaccurate. The true limit is
$\lim_{z\to 0} z \log z = 0$
All you have demonstrated is that, for a particular limiting sequence,
$0^0 = 1$

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