prove-0-0-2-

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

prove; (\frac{0}{0}) = 2

Commented by FilupS last updated on 20/Apr/17

\frac{0}{0} = undefined

Commented by FilupS last updated on 20/Apr/17

\lim_{x \to 0} \frac{x}{a} = 0 (∣ a ∣> 0)

\lim_{x \to 0} \frac{a}{x} = \pm \infty (∣ a ∣> 0)

by notation :

\lim_{L \to c} \frac{m}{n} = \frac{\lim_{L \to c} m}{\lim_{L \to c} n}

or

\lim_{L \to c} m n = (\lim_{L \to c} m) (\lim_{L \to c} n)

∴ \lim_{x \to 0} \frac{x}{x} = (\lim_{x \to 0} x) (\lim_{x \to 0} \frac{1}{x}) = 0 \times \infty = undefined

because :

\lim_{x \to c} \frac{x}{x} = 1 (\forall x \neq 0)

if true for x = 0, then 1 = 0 \times \infty

if 0 \times \infty = 0 \Rightarrow 1 = 0

if 0 \times \infty = \infty \Rightarrow 1 = \infty

both statements make no sense

hence \frac{0}{0} = undefined

Edit :

this proof holds true when : \frac{d}{d m} m = \frac{d}{d n} n = 0

Commented by mrW1 last updated on 20/Apr/17

\lim_{x \to 0} (2 \sin x) = 0

\lim_{x \to 0} (x) = 0

b u t \lim_{x \to 0} \frac{2 \sin x}{x} = 2

Commented by FilupS last updated on 20/Apr/17

interesting

Commented by chux last updated on 20/Apr/17

i love this

Commented by geovane10math last updated on 20/Apr/17

(\begin{matrix} 0 \\ 0 \end{matrix}) = 2

\frac{0!}{0! (0 - 0)!} = \frac{1}{1 \cdot 1} = 1 \neq 2

Commented by FilupS last updated on 20/Apr/17

i added an edit

Answered by chux last updated on 20/Apr/17

\frac{0}{0} = \frac{100 - 100}{100 - 100} = \frac{10^{2} - 10^{2}}{10 (10 - 10)}

= \frac{(10 - 10) (10 + 10)}{10 (10 - 10)}

divide common terms

= \frac{10 + 10}{10} = 20 / 10 = 2

note if the nethod is reversed we

would have 1 / 2

that \frac{0}{0} = 2 is a mathematical

falacy

Commented by Joel576 last updated on 20/Apr/17

you cant divide a number by 0

Commented by FilupS last updated on 20/Apr/17

\frac{(10 - 10) (10 + 10)}{10 (10 - 10)} = \frac{10 - 10}{10 - 10} \times \frac{10 + 10}{10}

= \frac{0}{0} \times 2

Commented by chux last updated on 20/Apr/17

thats true \dots . thanx for the correction .

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

thanks a lot

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

\frac{0}{0} = \lim_{x \to 1} \frac{x^{2} - 1}{x - 1} = \lim_{x \to 1} \frac{(x + 1) (x - 1)}{x - 1} =

\lim_{x \to 1} (x + 1) = 2 .

w e h a v e a f a t a l e r r o r t h a t d i v i d i n g b y

(x - 1) t h a t e q u a i l s t o z e r o i n \lim .

Commented by FilupS last updated on 21/Apr/17

this is incorrect

\lim_{x \to 1} \frac{(x + 1) (x - 1)}{x - 1} \neq 2

\lim_{x \to 1} \frac{(x + 1) (x - 1)}{x - 1} = \lim_{x \to 1} \frac{x - 1}{x - 1} (x + 1)

= \frac{0}{0} (2) = undefined

Commented by mrW1 last updated on 21/Apr/17

b u t i t i s c o r r e c t!

b y l i m i t w i t h x \to 1 i t m e a n s x i s n e a r 1 b u t \neq 1, a n d x - 1 \neq 0,

s o y o u c a n d i v i d e w i t h x - 1 .

\lim_{x \to 1} \frac{(x + 1) (x - 1)}{x - 1} = \lim_{x \to 1} (x + 1) = 2

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