# Proof of lim(1/x) x->0 by negating epsilon delta definition of limit

1. Oct 6, 2012

### can93

1. The problem statement, all variables and given/known data
I want to show that $\lim_{x \rightarrow 0}\frac{1}{x}$ does not exist by negating epsilon-delta definition of limit.

2. Relevant equations

3. The attempt at a solution
We say limit exists when:
$\forall \epsilon > 0, \exists \delta > 0 : \forall x(0< \left| x\right| < \delta \Rightarrow \left| \frac{1}{x} - L \right| < \epsilon)$
And limit does not exist corresponds to:
$\exists \epsilon > 0, \forall \delta > 0 : \exists x(0< \left| x\right| < \delta \wedge \left| \frac{1}{x} - L \right| > \epsilon)$ (for any L ?)

We look for an \epsilon such that

$\left| \frac{1}{x} - L \right| > \epsilon$

then using triangle inequality,

$\left| \frac{1}{x}\right| + \left| L \right| ≥ \left| \frac{1}{x} - L \right| > \epsilon$

therefore

$\left| \frac{1}{x}\right| + \left| L \right| > \epsilon$

and

$| \frac{1}{x}| > \epsilon - \left| L \right|$

for $\epsilon > |L|$, ( the other possibility gives no info since |x| > 0 already)

$|x| < \frac{1}{\epsilon - |L|}$

$0<|x|<\delta$

hence,

$|x| < min( \delta , \frac{1}{\epsilon - |L|} )$

we found such x satisfies the above condition, for some epsilon and L.

Is that approach correct?? Especially, I am not really sure about the negation of the definition...

Last edited: Oct 6, 2012
2. Oct 6, 2012

### Zondrina

Ah now you have to make some alterations to the definition for it to work properly. You have a limit involving an infinite quantity which requires a change in your definition to be :

$\forall ε>0, \exists δ>0 | 0<|x-a|<δ \Rightarrow f(x)>ε$

So for f(x) = 1/x, you desire a δ which ensures that f will be bigger than epsilon no matter what. You can start finding this delta by massaging the expression f(x) > ε to suit your needs.

3. Oct 6, 2012

### can93

alright,

for limits not dealing with infinity, is my negation correct??

4. Oct 6, 2012

### Zondrina

I'm not sure why you would want to negate the statement because |f(x)-L| which translates into |f(x)-∞| is for lack of better term, a garbage statement.

That's why I suggested using the modified form of the definition so it at least makes sense.