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Question Number 12332 by frank ntulah last updated on 19/Apr/17

prove ; ((0/0))=2

prove;(00)=2

Commented by FilupS last updated on 20/Apr/17

(0/0)=undefined

00=undefined

Commented by FilupS last updated on 20/Apr/17

lim_(x→0) (x/a)=0      (∣a∣>0)  lim_(x→0) (a/x)=±∞      (∣a∣>0)     by notation:  lim_(L→c) (m/n)=((lim_(L→c) m)/(lim_(L→c) n))  or  lim_(L→c) mn=(lim_(L→c) m)(lim_(L→c) n)    ∴lim_(x→0) (x/x)=(lim_(x→0) x)(lim_(x→0) (1/x))=0×∞=undefined  because:  lim_(x→c) (x/x)=1   (∀x≠0)  if true for x=0,  then    1=0×∞  if   0×∞=0   ⇒  1=0  if   0×∞=∞ ⇒  1=∞    both statements make no sense  hence (0/0)=undefined     Edit:  this proof holds true when:  (d/dm)m=(d/dn)n=0

limx0xa=0(a∣>0)limx0ax=±(a∣>0)bynotation:limLcmn=limLcmlimLcnorlimLcmn=(limLcm)(limLcn)limx0xx=(limx0x)(limx01x)=0×=undefinedbecause:limxcxx=1(x0)iftrueforx=0,then1=0×if0×=01=0if0×=1=bothstatementsmakenosensehence00=undefinedEdit:thisproofholdstruewhen:ddmm=ddnn=0

Commented by mrW1 last updated on 20/Apr/17

lim_(x→0)  (2sin x)=0  lim_(x→0)  (x)=0  but lim_(x→0)  ((2sin x)/x)=2

limx0(2sinx)=0limx0(x)=0butlimx02sinxx=2

Commented by FilupS last updated on 20/Apr/17

interesting

interesting

Commented by chux last updated on 20/Apr/17

i love this

ilovethis

Commented by geovane10math last updated on 20/Apr/17

  ((0),(0) ) = 2  ((0!)/(0!(0 − 0)!)) = (1/(1∙1)) = 1 ≠ 2

(00)=20!0!(00)!=111=12

Commented by FilupS last updated on 20/Apr/17

i added an edit

iaddedanedit

Answered by chux last updated on 20/Apr/17

(0/0)=((100−100)/(100−100))=((10^2 −10^2 )/(10(10−10)))  =(((10−10)(10+10))/(10(10−10)))      divide common terms    =((10+10)/(10))=20/10 =2      note if the nethod is reversed we   would have 1/2    that (0/0)=2 is a mathematical  falacy

00=100100100100=10210210(1010)=(1010)(10+10)10(1010)dividecommonterms=10+1010=20/10=2noteifthenethodisreversedwewouldhave1/2that00=2isamathematicalfalacy

Commented by Joel576 last updated on 20/Apr/17

you cant divide a number by 0

youcantdivideanumberby0

Commented by FilupS last updated on 20/Apr/17

(((10−10)(10+10))/(10(10−10)))=((10−10)/(10−10))×((10+10)/(10))  =(0/0)×2

(1010)(10+10)10(1010)=10101010×10+1010=00×2

Commented by chux last updated on 20/Apr/17

thats true.... thanx for the correction.

thatstrue....thanxforthecorrection.

Commented by frank ntulah last updated on 24/Apr/17

thanks a lot

thanksalot

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Apr/17

(0/0)=lim_(x→1) ((x^2 −1)/(x−1))=lim_(x→1) (((x+1)(x−1))/(x−1))=  lim_(x→1) (x+1)=2   .■  we have a fatal error that dividing by  (x−1)that equails to zero in lim.

00=limx1x21x1=limx1(x+1)(x1)x1=limx1(x+1)=2.wehaveafatalerrorthatdividingby(x1)thatequailstozeroinlim.

Commented by FilupS last updated on 21/Apr/17

this is incorrect  lim_(x→1) (((x+1)(x−1))/(x−1))≠2     lim_(x→1) (((x+1)(x−1))/(x−1))=lim_(x→1) ((x−1)/(x−1))(x+1)  =(0/0)(2)=undefined

thisisincorrectlimx1(x+1)(x1)x12limx1(x+1)(x1)x1=limx1x1x1(x+1)=00(2)=undefined

Commented by mrW1 last updated on 21/Apr/17

but it is correct!  by limit with x→1 it means x is near 1 but ≠1, and x−1≠0,   so you can divide with x−1.  lim_(x→1) (((x+1)(x−1))/(x−1))=lim_(x→1)  (x+1)=2

butitiscorrect!bylimitwithx1itmeansxisnear1but1,andx10,soyoucandividewithx1.limx1(x+1)(x1)x1=limx1(x+1)=2

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