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

[can93]

Homework Statement

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

Homework Equations

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|}

we also had

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...

 

[STEMucator]

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.

 

[can93]

alright,

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

 

[STEMucator]

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.

To answer your question though, I don't see a problem with your negation.