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

## Homework Statement

I want to show that $\lim_{x \rightarrow 0}\frac{1}{x}$ does not exist by negating epsilon-delta definition of limit.

## 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...

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STEMucator
Homework Helper
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.

alright,

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

STEMucator
Homework Helper
alright,

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