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/sci/ - Science & Math

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Anonymous Mon Jun 28 23:15:59 2010 No.1292402 [Reply] [Original]

Can somebody shed some light on this?

>>	Anonymous Mon Jun 28 23:17:59 2010 No.1292420 >>1292402 what about sign(x)

>>	Anonymous Mon Jun 28 23:18:59 2010 No.1292424 wtf is that possible

>>	Anonymous Mon Jun 28 23:19:59 2010 No.1292429 use standard differentiation

>>	Anonymous Mon Jun 28 23:21:59 2010 No.1292439 >>1292420 thanks

>>	Anonymous Mon Jun 28 23:22:59 2010 No.1292446 thats easy i'll give you one, x^3

>>	Anonymous Mon Jun 28 23:22:59 2010 No.1292448 \|x\|

>>	Anonymous Mon Jun 28 23:23:59 2010 No.1292451 >>1292446 wrong >>1292448 wrong

>>	Anonymous Mon Jun 28 23:23:59 2010 No.1292455 arctan(x)

>>	Anonymous Mon Jun 28 23:24:59 2010 No.1292465 >>1292455 Also wrong. Especially since arctan(0) is easily evaluated. It's 0.

>>	Anonymous Mon Jun 28 23:27:59 2010 No.1292477 >>1292465 limits have nothing to do with the function's value.

>>	Anonymous Mon Jun 28 23:28:59 2010 No.1292485 >>1292455 the limit on arctan(x) and arctan(x^2) are both 0.

>>	Anonymous Mon Jun 28 23:29:59 2010 No.1292493 >>1292489 He's right but in this case it doesn't matter because it's continuous anyway.

>>	Anonymous Mon Jun 28 23:29:59 2010 No.1292492 >>1292477 Yes, yes they do. The question was, find a function for which the limit as x approaches zero of f(x) cannot be evaluated, but f(x^2) can. Limit as x approaches zero of arctan(x) is easily evaluated.

>>	Anonymous Mon Jun 28 23:29:59 2010 No.1292490 f(x) = x for x>= 0 f(x) = sqrt(x) for x<0 They don't approach same value from left as right but square will

>>	Anonymous Mon Jun 28 23:29:59 2010 No.1292489 >>1292477 True, but arctan(x) is smooth.

>>	Anonymous Mon Jun 28 23:30:59 2010 No.1292496 >>1292490 x<0 for sqrt(x) is not part of the domain. lim sqrt(x) x->0 is actually 0.

>>	Anonymous Mon Jun 28 23:37:59 2010 No.1292541 f(x) = 1/((dy/dx) /2x) did i win?

>>	Anonymous Mon Jun 28 23:38:59 2010 No.1292545 and as usual no one notices the simplest function that works was mentioned first...and no one notices

>>	Anonymous Mon Jun 28 23:38:59 2010 No.1292546 >>1292451 How is >>1292448 wrong? f(x) = \|x\|; limit doesn't exist f(x^2) = \|x^2\| = x^2; limit exists

>>	Anonymous Mon Jun 28 23:38:59 2010 No.1292547 >>1292541 I won!!! haza!

>>	Anonymous Mon Jun 28 23:39:59 2010 No.1292551 >>1292545 sign(x) is to boring, we need a little spice

>>	Anonymous Mon Jun 28 23:40:59 2010 No.1292555 f(x) = x^-1

>>	Anonymous Mon Jun 28 23:40:59 2010 No.1292556 >>1292541 f(x) = 1/((dy/dx) /sqrt(2x))

>>	Anonymous Mon Jun 28 23:40:59 2010 No.1292557 >>1292546 The limit of f(x) = \|x\| exists just fine. Its zero at zero. Its even continuous.

>>	Anonymous Mon Jun 28 23:40:59 2010 No.1292558 File: 39 KB, 469x428, trollface.jpg [View same] [iqdb] [saucenao] [google] >>1292546 http://www.wolframalpha.com/input/?i=lim+\|x\|+x-%3E0

>>	Anonymous Mon Jun 28 23:41:59 2010 No.1292561 >>1292555 wrong, a limit being infinity isn't a limit.

>>	Anonymous Mon Jun 28 23:42:59 2010 No.1292566 >>1292557 sorry my bad, I'm an idiot ... got limits and derivatives mixed up :(

>>	Anonymous Mon Jun 28 23:42:59 2010 No.1292567 In some sense, sqrt(x) works. As itself, its a non analytic function, there is no direction independent limit. while f(x^2) is entire.

>>	Anonymous Mon Jun 28 23:43:59 2010 No.1292570 f(x) = -1 if x<0 f(x) = 1 if x>0

>>	Anonymous Mon Jun 28 23:43:59 2010 No.1292571 >>1292546 >f(x) = \|x\|; limit doesn't exist What the fuck? It's continuous on the reals.

>>	Anonymous Mon Jun 28 23:43:59 2010 No.1292573 >>1292546 <span class="math"> \lim_{x \to 0} \|x\| = 0 [/spoiler] It isn't continuous at 0 but the limit of f exists.

>>	Anonymous Mon Jun 28 23:44:59 2010 No.1292575 x-1 for x<=0 x+1 for x>0 Squared both approach 1 Normal approach -1 or 1

>>	Anonymous Mon Jun 28 23:44:59 2010 No.1292576 >>1292570 what is this i dont even

>>	Anonymous Mon Jun 28 23:44:59 2010 No.1292577 >>1292573 *differentiable

>>	Anonymous Mon Jun 28 23:44:59 2010 No.1292578 >>1292576 equal to sign(x) actually.

>>	Anonymous Mon Jun 28 23:44:59 2010 No.1292579 >>1292576 14 year old detected.

>>	Anonymous Mon Jun 28 23:45:59 2010 No.1292581 itt: retards who dont understand limits

>>	Anonymous Mon Jun 28 23:45:59 2010 No.1292582 >>1292573 Then \|x\|/x. Hooha! (Not other poster.)

>>	Anonymous Mon Jun 28 23:46:59 2010 No.1292587 >>1292448 \|x\| Is correct. And obvious.

>>	Anonymous Mon Jun 28 23:46:59 2010 No.1292589 >>1292570 = >>1292420 >>1292545 is right

>>	Anonymous Mon Jun 28 23:47:59 2010 No.1292591 >>1292582 \|x\|/x = sign(x) = first reply in this thread

>>	Anonymous Mon Jun 28 23:49:59 2010 No.1292595 >>1292587 is troll. is obvious

>>	Anonymous Mon Jun 28 23:49:59 2010 No.1292596 >>1292589 They aren't equal. Sign(x) is -1 for x less than zero and 1 for x greater.

>>	Anonymous Mon Jun 28 23:50:59 2010 No.1292600 >>1292596 yes, and >>1292591 is... ?

>>	Anonymous Mon Jun 28 23:51:59 2010 No.1292603 >>1292596 > they aren't equal > states definition that shows they are equal wut

>>	Anonymous Mon Jun 28 23:54:59 2010 No.1292617 > PhD in Math > 300k starting > any version of sgn(x) I want

>>	Anonymous Mon Jun 28 23:54:59 2010 No.1292620 File: 34 KB, 462x477, 1277057984976.jpg [View same] [iqdb] [saucenao] [google] >this thread

>>	Anonymous Mon Jun 28 23:59:51 2010 No.1292645 f(x)=x/sin(x^2)

>>	Anonymous Tue Jun 29 00:00:54 2010 No.1292653 >>1292645 Oops... meant x/sin^2(x)

>>	Anonymous Tue Jun 29 00:02:42 2010 No.1292664 >>1292645 >implying the question wasnt answered in the first reply

>>	Anonymous Tue Jun 29 00:15:40 2010 No.1292739 limits are gay calculs is gay algebra is the msot general theory of matematics

>>	Anonymous Tue Jun 29 00:27:51 2010 No.1292798 >>1292739 lol apparently you believe spelling is gay as well

>>	Anonymous Tue Jun 29 00:39:25 2010 No.1292859 >>1292561 Nah bro pretty sure that exists.