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## Homework Statement

Prove that [itex]\lim_{x\rightarrow 0} \frac{1}{x}[/itex] does not exist using the precise definitions of a limit.

## Homework Equations

Precise laws for infinite limits:

[itex]\lim_{x\rightarrow a} f(x) = \infty[/itex] means that for every positive number M there is a positive number [itex]\delta[/itex] such that if [itex]0 < |x-a| < \delta[/itex] then [itex]f(x)>M[/itex]

and

[itex]\lim_{x\rightarrow a} f(x) = - \infty[/itex] means that for every negative number N there is a positive number [itex]\delta[/itex] such that if [itex]0 < |x-a| < \delta[/itex] then [itex]f(x) < N[/itex]

There's also the precise laws for LHS- and RHS-limits:

[itex]\lim_{x\rightarrow a^-} f(x) = L[/itex] if for every number [itex]\epsilon > 0[/itex] there is a number [itex]\delta > 0[/itex] such that if [itex]a - \delta < x < a[/itex] then [itex]|f(x) - L| < \epsilon[/itex]

and

[itex]\lim_{x\rightarrow a^+} f(x) = L[/itex] if for every number [itex]\epsilon > 0[/itex] there is a number [itex]\delta > 0[/itex] such that if [itex]a < x < a + \delta[/itex] then [itex]|f(x) - L| < \epsilon[/itex]

And, of course:

If [itex]\lim_{x\rightarrow a^-} f(x) = L[/itex] and [itex]\lim_{x\rightarrow a^+} f(x) = L[/itex], then [itex]\lim_{x\rightarrow a} f(x) = L[/itex]. Otherwise, it is undefined.

## The Attempt at a Solution

Well, I think I've successfully proved it, but I'm not completely sure. My idea was to use the RHS- and LHS-limits to show that they differed, which would then show that the original limit does not exist. Here's my proof:

Proof of [itex]\lim_{x\rightarrow 0^+} \frac{1}{x} = \infty[/itex]:

We have to define delta such that if [itex]0 < x < \delta[/itex], then [itex]\frac{1}{x} > M[/itex]

[itex]\frac{1}{x} > M \rightarrow x < \frac{1}{M}[/itex]

Choose [itex]\delta = \frac{1}{M} \rightarrow 0 < x < \delta \rightarrow \frac{1}{x} > M[/itex]

This shows that [itex]\lim_{x \rightarrow 0^+} \frac{1}{x} = \infty[/itex] per the precise definition of a limit.

Proof of [itex]\lim_{x\rightarrow 0^-} \frac{1}{x} = - \infty[/itex]:

We have to define delta such that if [itex]- \delta < x < 0[/itex], then [itex]\frac{1}{x} < N[/itex]

[itex]\frac{1}{x} < N \rightarrow x > \frac{1}{N}[/itex]

Choose [itex]\delta = \frac{-1}{N} \rightarrow - \delta < x < 0 \rightarrow \frac{-1}{N} < N[/itex]

This shows that [itex]\lim_{x \rightarrow 0^-} \frac{1}{x} = - \infty[/itex] per the precise definition of a limit.

Conclusion:

[itex]\lim_{x \rightarrow 0^+} \frac{1}{x} \neq \lim_{x \rightarrow 0^-} \frac{1}{x}[/itex]

Therefore, the original limit does not exist.

...At least, that's the idea. I'm very unfamiliar with proofs, so did I do anything wrong? Would this be considered adequate proof?

EDIT: I just read that what my book calls the 'precise definition of a limit' is more officially known as the epsilon-delta limit definition.

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