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

1. Dec 31, 2008

### farleyknight

1. The problem statement, all variables and given/known data

Prove that $\lim_{x \to 1} \frac{x}{x - 1} = \infty$

2. Relevant equations

3. The attempt at a solution

First $0 < |x - 1| < \delta$ implies $\frac{x}{x - 1} > N, \forall N$

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

then $\frac{n}{x - 1} > N$
$\frac{1}{x - 1} > \frac{N}{n}$
$x - 1 < \frac{n}{N}$

So choose $\delta = \frac{n}{N}$

and $0 < |x - 1| < \delta$ 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

2. Dec 31, 2008

### Dick

No. Not so good. For one thing the limit is only infinity if x>1 (x approaching 1 from the right). If x approaches from the left it's negative infinity. Start by taking your condition that x/(x-1)>N. Change that into an inequality condition for x. Can you deduce from that how to pick a delta?