(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove that [itex]\lim_{x \to 1} \frac{x}{x - 1} = \infty[/itex]

2. Relevant equations

3. The attempt at a solution

First [itex]0 < |x - 1| < \delta[/itex] implies [itex]\frac{x}{x - 1} > N, \forall N[/itex]

Because the reals are dense, we can choose an [itex]n > 0[/itex] such that [itex]\frac{x}{x - 1} > \frac{n}{x - 1} > N[/itex]

then [itex]\frac{n}{x - 1} > N[/itex]

[itex]\frac{1}{x - 1} > \frac{N}{n}[/itex]

[itex]x - 1 < \frac{n}{N}[/itex]

So choose [itex]\delta = \frac{n}{N}[/itex]

and [itex]0 < |x - 1| < \delta[/itex] holds

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My first question is of course, is this proof correct. Secondly, if it's not, my other question is, is choosing an n as I've done above, a legal operation? Could it be changed to make it a legal operation?

Thanks,

- Rob

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# Lim {x -> 1} x / (x - 1) = infty

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