# Limits involving e and ln

1. Mar 3, 2012

### Tomath

1. The problem statement, all variables and given/known data
Show that lim (x,y) → (a,0) e^(x ln y) = 0 $\forall$a > 0

2. Relevant equations

3. The attempt at a solution
I've tried looking at lim (x,y) → (a,0) x ln y seperately.
lim(x,y) → (a,0) x ln y = lim(x,y) -> (a,0) x * lim(x,y) → (a,0) ln y
= a * lim(x,y) → (a,0) ln y

Now lim(x,y) → (a,0) ln y is -∞. So we get lim(x,y) →(a,0) x ln y = -∞. Now lim z → -∞ e^z = 0.

The problem here is that we haven't discussed limits involving infinity in class and i'm pretty sure i'm not allowed to use it. My question therefore is: is there any other way to show that lim (x,y) → (a,0) e^(x ln y) = 0?

2. Mar 3, 2012

### Dick

No, I think that's the right way to do it.

3. Mar 3, 2012

### Tomath

That's what I was afraid off :(. How would I prove that lim (x,y) → (a,0) e^(x ln y) = 0 and lim z → -∞ e^z = 0? Im familar with the $\delta$ and $\epsilon$ method but im not sure how to include ∞ in this method.

4. Mar 3, 2012

### Dick

lim z → -∞ e^z = 0 just means that for any ε>0 you can find a N such that e^z<ε if z<N. Try picking N=log(ε).