# X^x=e^xlnx continuity

1. Nov 15, 2008

### 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?

2. Nov 15, 2008

### Office_Shredder

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

3. Nov 15, 2008

### kathrynag

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

4. Nov 15, 2008

### 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.

5. Nov 15, 2008

### kathrynag

Ok, how do I do that epsilon delta proof?

6. Nov 15, 2008

### 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<$$\epsilon$$

7. Nov 15, 2008

### 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<$$\epsilon$$.

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

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<$$\epsilon$$?

8. Nov 15, 2008

### kathrynag

Should it be changed to a f or are you talking about a change from x to y?

Ok so there is d' such that lg(x)-g(a)l<$$\delta$$. then |f(g(x))-f(g(a)l<$$\epsilon$$

9. Nov 19, 2008

### 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 $$\left|y-x_{0}g(x_{0})\right|$$<$$\delta_{1}$$.

Last edited: Nov 19, 2008