# function doesn't tend to -infty or infty, how to prove it diverges?

• Mar 1st 2011, 01:42 AM
mremwo
function doesn't tend to -infty or infty, how to prove it diverges?
How would I show that lim as x goes to 1 of x/(x-1) does not exist, using a rigorous proof?

I am a little confused since this function neither tends to -infinity or infinity (at least not that I think it does). I started off by proving this fact, but it turns out that I don't think it is getting me anywhere. Any suggestions?

Thanks.
• Mar 1st 2011, 01:58 AM
Debsta
Might help if you write the function as 1 + 1/(x-1) and go from there.
• Mar 1st 2011, 03:04 AM
HallsofIvy
Quote:

Originally Posted by mremwo
How would I show that lim as x goes to 1 of x/(x-1) does not exist, using a rigorous proof?

I am a little confused since this function neither tends to -infinity or infinity (at least not that I think it does). I started off by proving this fact, but it turns out that I don't think it is getting me anywhere. Any suggestions?

Thanks.

It tends to +infinity as x goes to 1 from above, -infinity as x goes to 1 from below. Either is enough to show that it does not converge. Remember that, to prove a general statement is not true, a counter example is sufficient. Take $\displaystyle \epsilon= 1$, for simplicity. If x> 1 then let $\displaystyle \delta= x- 1> 0$. $\displaystyle x= 1+ \delta$ so $\displaystyle \left|\frac{x}{x- 1}- 1\right|= \left|\frac{1+ \delta}{\delta}- 1\right|= \frac{1}{\delta}$ (notice how debsta's "1+ 1/(x-1)" came into that) which will be larger that $\displaystyle \epsilon= 1$ as long as [itex]\delta< 1[/tex]. We cannot make it small by taking $\displaystyle \delta$ smaller.