Question

How is 0=0/x, considering x can be 0? http://www.wolframalpha.com/input/?i=0%3D0%2Fx

Answer 1

0=0/x is true for all real numbers except x=0

Answer 2

0/0 is undefined

Answer 3

So how can it be said that the statement is always true?

Answer 4

I ask because, in my calculus book, it gives a proof that the derivative of a constant function is 0 (it uses 0/something = 0).

Answer 5

because "always" is a relative term.

can u say that 2+2=4 always?

Answer 6

u have
\[ \large \lim_{h\to0}\frac{0}{h}=\lim_{h\to0}0=0 \]
not
\[ \large =\frac{0}{0} \]

Answer 7

That is saying 0/h=0. That isn't always true, so how does the proof work?

Answer 8

because \(h\to0\) but \(h\neq0\)

Answer 9

Why is \[h \neq0\]?

Answer 10

because of the definition of limit
\[ \large \lim_{x\to a}f(x)=L \Leftrightarrow \]
\[ \large (\forall\varepsilon>0)(\exists\delta_\varepsilon>0)(\forall x)
(0<|x-a|<\delta_\varepsilon\Rightarrow|f(x)-L|<\varepsilon) \]

Answer 11

do u notice the \(0<|x-a|\) in the definition?
what does it mean?

Answer 12

The denominator cannot be 0.

Answer 13

in your example, yes

Answer 14

Thanks, but I still have issue with 0=0/x.

Answer 15

(Outside of the calculus problem.)

Answer 16

u r welcome