/sci/ - Science & Math » Thread #13216120

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What’s wrong with this proof?

Anonymous Mon 31 May 11:01:40 2021 No.13216120 View Reply Original Report

Quoted By: >>13216139 >>13216145 >>13216156 >>13216169 >>13216223

Anonymous

Anonymous Mon 31 May 2021 11:03:46 No.13216124 Report

Quoted By:

1*0=2*0
Therefore 1=2

Anonymous

Anonymous Mon 31 May 2021 11:08:34 No.13216139 Report

Quoted By:

>>13216120
The main problem is that it's not correct.

Anonymous

Anonymous Mon 31 May 2021 11:10:54 No.13216145 Report

Quoted By:

>>13216120
You sure line 3 is a valid step? I don't think you can multiply through by lim 1/h.

Anonymous

Anonymous Mon 31 May 2021 11:13:16 No.13216156 Report

Quoted By: >>13216173 >>13216203

>>13216120
The part where h in the denominator miraculously disappears. Probably taken out of the limit to multiply 1*h on the right side and you obviously can't take the limiting variable outside of its limit.

Anonymous

Anonymous Mon 31 May 2021 11:15:30 No.13216169 Report

Quoted By: >>13216262

>>13216120
the implication doesn't go the other way around, no?

Anonymous

Anonymous Mon 31 May 2021 11:16:31 No.13216173 Report

Quoted By:

>>13216156
also this

Anonymous

Anonymous Mon 31 May 2021 11:26:10 No.13216203 Report

Quoted By: >>13216207

>>13216156
He said that the h disappears on the bottom because in order for the limit to equal 1 the top part has to equal 0

Anonymous

Anonymous Mon 31 May 2021 11:29:10 No.13216207 Report

Quoted By: >>13216220

>>13216203
classic 0/0 bullshit then

Anonymous

Anonymous Mon 31 May 2021 11:33:05 No.13216220 Report

Quoted By:

>>13216207
But you can do limits that are 0/0

Anonymous

Anonymous Mon 31 May 2021 11:34:48 No.13216223 Report

Quoted By: >>13216262

>>13216120
$\lim_{h\to0}\frac{f(x+h)}{f(x)}-1 = 0$ does not imply $\lim_{h\to0}\frac{\frac{f(x+h)}{f(x)}-1}{h} = 1$
In this case $\lim_{h\to0}\frac{(x+h)}{(x)}-1 = 0$
but $\lim_{h\to0}\frac{\frac{(x+h)}{(x)}-1}{h} = \lim_{h\to0}\frac{x+h-x}{xh} = \lim_{h\to0}\frac{h}{xh} = \frac{1}{x}$
The mistake is assuming that f(x) satisfying (**) implies f(x) satisfying (*).

Anonymous

Anonymous Mon 31 May 2021 11:51:36 No.13216262 Report

Quoted By:

>>13216169
>>13216223
seems correct

Anonymous

Anonymous Mon 31 May 2021 17:59:34 No.13217638 Report

Quoted By:

1 = 1
so what?

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