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can93
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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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