Trying to prove the continuity of f(x) = ln (x) on (0, infin). Sorry for the terrible formatting. Thanks for any help.

So, given a in (0, infin), and given epsilon > 0, let delta = min (e^-epsilon, e^ epsilon).

Then abs(x-a) < delta implies that e^-epsilon < x-a < e^epsilon

this implies e^-epsilon < e^ln(x)-e^ln(a) < e^epsilon

which implies -epsilon < ln(x)-ln(a) < epsilon

thus, abs(ln(x)-ln(a)) < epsilon as desired.

Thus f is continuous at a.

let delta = min (e^-epsilon, e^ epsilon)

"Then abs(x-a) < delta implies that e^-epsilon < x-a < e^epsilon"

This step confuses me. Doesn't this just mean that abs(x-a) < min(e^-epsilon, e^ epsilon)
so x-a < e^-epsilon and x-a < e^epsilon

Maybe I'm misundestanding your notation (the min, is that minimum or what?)

also

"Then abs(x-a) < delta implies that e^-epsilon < x-a < e^epsilon

this implies e^-epsilon < e^ln(x)-e^ln(a) < e^epsilon"

ln(x-a) is not the same as ln(x) - ln(a) (that would be ln(x/a)!)

So the best you could do with x-a is write it as e^ln(x-a)



I'll write a proof of this if you'd like but atm a lil buys with work lol

My proof (feel free to copy, point out errors, ask questions):

say a,b in (0,infinity), |a-b| < delta, where delta > 0

|ln(a) - ln(b)| = |ln(a/b)| = |ln((a-b)/b+1)| < |ln(delta/b+1)|

so what kind of delta satisfies |ln(delta/b+1)| < epsilon ?

since delta > 0 (and the ln expresion can't be < 0), we know this is the same as

ln(delta/b+1) < epsilon

so delta < b((e^epsilon)-1)


so we can always find such a delta s.t |ln(a)-ln(b)| < epsilon
because when delta < b((e^epsilon)-1), then this |ln(delta/b+1)| < epsilon, and |ln(a) - ln(b)| < |ln(delta/b+1)|


Note: delta < b((e^epsilon)-1), positive deltas can satisfy this because b > 0 and epsilon > 0. Epsilon greater 0 confirms e^epsilon > 1, so b((e^epsilon)-1) > 0

Note2: so I did this for a-b < delta, forgot to also do a-b > -delta part. This should very much mirror what I have there already. If you have trouble with this part lmk

should be |ln((a-b)/b+1)| < |ln(delta/b+1)| or |ln((a-b)/b+1)| < |ln(-delta/b+1)|
then you prove |ln(-delta/b+1)| > -epsilon

This post was edited by Segfault on Sep 9 2013 10:58am

Looks solid.

Here's the rest of the proof if you still need it.

|ln(a) - ln(b)| = |ln(a/b)| = |ln((a-b)/b+1)|

|ln((a-b)/b+1)| <= |ln(delta/b+1)| or |ln((a-b)/b+1)| <= |ln(-delta/b+1)|

if |ln((a-b)/b+1)| < |ln(delta/b+1)| see segfault's proof (although replace some of the strict inequalities with >=
if |ln((a-b)/b+1)| < |ln(-delta/b+1)| then

|ln(-delta/b+1)| <= epsilon, which is the same as ln(-delta/b+1) <= epsilon

-delta <= b((e^epsilon)-1) or delta >= b((e^epsilon)-1)

so Segfault's proof showed that delta <= b((e^epsilon)-1)

So this shows that, for |f(a) - f(b)| < epsilon, we can find a delta such that |a-b| < delta, and this delta = b((e^epsilon)-1)