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## Homework Statement

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

## Homework Equations

## The Attempt at a Solution

We say limit exists when:

[itex]\forall \epsilon > 0, \exists \delta > 0 : \forall x(0< \left| x\right| < \delta \Rightarrow \left| \frac{1}{x} - L \right| < \epsilon) [/itex]

And limit does not exist corresponds to:

[itex]\exists \epsilon > 0, \forall \delta > 0 : \exists x(0< \left| x\right| < \delta \wedge \left| \frac{1}{x} - L \right| > \epsilon) [/itex] (for any L ?)

We look for an \epsilon such that

[itex]\left| \frac{1}{x} - L \right| > \epsilon [/itex]

then using triangle inequality,

[itex]\left| \frac{1}{x}\right| + \left| L \right| ≥ \left| \frac{1}{x} - L \right| > \epsilon [/itex]

therefore

[itex]\left| \frac{1}{x}\right| + \left| L \right| > \epsilon [/itex]

and

[itex]| \frac{1}{x}| > \epsilon - \left| L \right| [/itex]

for [itex] \epsilon > |L|[/itex], ( the other possibility gives no info since |x| > 0 already)

[itex]|x| < \frac{1}{\epsilon - |L|}[/itex]

we also had

[itex]0<|x|<\delta[/itex]

hence,

[itex]|x| < min( \delta , \frac{1}{\epsilon - |L|} )[/itex]

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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