How can we prove the continuity of ln x over (0, ∞)?

[evagelos]

how can we prove that lun x is continuous over (0, $$\infty$$ )?

Provided that we define : lun x =y <=> $$e^y =x$$?

 

[mathman]

lnx - lnu = ln(x/u). For any fixed x > 0, u->x => x/u -> 1 and ln1=0.

You can dress this proof up.

 

[evagelos]

lnx - lnu = ln(x/u). For any fixed x > 0, u->x => x/u -> 1 and ln1=0.

You can dress this proof up.

Can you elaborate a little more ? I do not how to start

 

[mathman]

I am not sure what you are given to start with. For example are you assuming e^y is continuous? My proof (I admit) is somewhat flawed. It needs continuity of ln(x) for x=1, which then implies continuity for all x.