# X^x=e^xlnx continuity

## Homework Statement

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

## 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?

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

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<$$\epsilon$$

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.

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?

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<$$\epsilon$$

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<$$\epsilon$$?

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<$$\epsilon$$

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

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