First we have: (a is the indepedent variable).
Let be a irrational number. Let be a rational sequence with .
We get a sequence of functions .
Now we are to prove that , so that .
There is a theorem saying that:
If 1: ,
2: uniformly converges at ,
3: converges at at least one point ,
then .
we know that for every x, ,so 3 is true.
1 is clearly true.
Now we only have to show 2.Let
let , it suffices to show that
Let , then
so
We have .
so ,the second inequality is true since is uniformly continous on .
So ,we are done..
So tough a proof...
Or, if you first prove that the derivative of ln(x) is 1/x, starting from , take the ln of both sides: ln(y)= r ln(x).
Now, differentiating both sides with respect to x, . Multiply both sides by y: .
That's easier than ynj's proof but requires first proving that the derivative of ln(x) is 1/x as well as the chain rule.