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

  • Thread starter 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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Answers and Replies

  • #2
STEMucator
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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 :

[itex]\forall ε>0, \exists δ>0 | 0<|x-a|<δ \Rightarrow f(x)>ε[/itex]

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
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alright,

for limits not dealing with infinity, is my negation correct??
 
  • #4
STEMucator
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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.
 

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