x^x=e^xlnx continuity

[kathrynag]

1. The problem statement, all variables and given/known data
f(x)=e^x
g(x)=lnx
h(x)=x^x=e^xlnx

If f and g are continous prove h(x) is continious for x>0
2. Relevant equations



3. The attempt at a solution
Ok this confuses me, because I would think that it wouldn't be too bad too do if h(x)=f(g(x)). Maybe the book had a typo?

[Office_Shredder]

this is just f(x*g(x)). What do you know about x*g(x) if g is continuous?

[kathrynag]

this is just f(x*g(x)). What do you know about x*g(x) if g is continuous?

Then lx-x0l<\delta
Then lxg(x)-x_{0}g(x_{0})l<\epsilon

[snipez90]

Hmm well if you've proved that the product of two continuous functions is continuous (which is easy, provided that you know the proof that the limit of a product is the product of limits), then I don't think you need to resort to epsilon delta for x*g(x).

Also, f is continuous, so basically you have a lot of continuous functions involved in different operations. If you've already proved that f(g(x)) is continuous at a point a if g(x) is continuous at a and f is continuous at g(a), then you could save yourself a lot of trouble.

[kathrynag]

Also, f is continuous, so basically you have a lot of continuous functions involved in different operations. If you've already proved that f(g(x)) is continuous at a point a if g(x) is continuous at a and f is continuous at g(a), then you could save yourself a lot of trouble.

Ok, how do I do that epsilon delta proof?

[kathrynag]

lx-al<\delta
lg(x)-g(a)l<\epsilon

f continuous at g(a)
lx-g(a)l<\delta
lf(x)-f(g(a)l<\epsilon

[snipez90]

A few comments to point you in the right direction. We want |x-a|<\delta to imply that lf(g(x))-f(g(a)l<\epsilon.

lx-al<\delta
lg(x)-g(a)l<\epsilon


This is fine, except one variable should be changed. We want to somehow connect this to our second fact, which is

f continuous at g(a)
lx-g(a)l<\delta
lf(x)-f(g(a)l<\epsilon


Ok, obviously we want to be careful about using the same variables. Since x is used to represent elements in the domain of g, let's use y to represent the elements in the domain f. Making this change, we have

f continuous at g(a)
ly-g(a)l<\delta
lf(y)-f(g(a)l<\epsilon

Now note that if we replace y with g(x), which is certainly allowed, we are very close to what we need. Namely, lg(x)-g(a)l<\delta implies that |f(g(x))-f(g(a)l<\epsilon. But we also know that we reserved the variable delta for lx-al<\delta. So we need to change the delta in lg(x)-g(a)l<\delta to some other variable.

In fact, we can choose any variable that is greater than 0 (Why?). So let's choose d' (delta prime). Now how can we use the fact that d' > 0 to connect our delta in |x-a| < \delta to our epsilon in |f(g(x))-f(g(a)l<\epsilon?

[kathrynag]

A few comments to point you in the right direction. We want |x-a|<\delta to imply that lf(g(x))-f(g(a)l<\epsilon.



This is fine, except one variable should be changed. We want to somehow connect this to our second fact, which is
Should it be changed to a f or are you talking about a change from x to y?



Ok, obviously we want to be careful about using the same variables. Since x is used to represent elements in the domain of g, let's use y to represent the elements in the domain f. Making this change, we have

f continuous at g(a)
ly-g(a)l<\delta
lf(y)-f(g(a)l<\epsilon

Now note that if we replace y with g(x), which is certainly allowed, we are very close to what we need. Namely, lg(x)-g(a)l<\delta implies that |f(g(x))-f(g(a)l<\epsilon. But we also know that we reserved the variable delta for lx-al<\delta. So we need to change the delta in lg(x)-g(a)l<\delta to some other variable.

In fact, we can choose any variable that is greater than 0 (Why?). So let's choose d' (delta prime). Now how can we use the fact that d' > 0 to connect our delta in |x-a| < \delta to our epsilon in |f(g(x))-f(g(a)l<\epsilon?
Ok so there is d' such that lg(x)-g(a)l<\delta. then |f(g(x))-f(g(a)l<\epsilon

[kathrynag]

Ok, here's my start:
h(x)=x^{x}
Then h(x)=f(xg(x))
Choose \epsilon<0. There is \delta_{1}>0 such that if \left|y-x_{0}g(x_{0})\right|<\delta_{1}.