Indeterminant forms HELP

does an indeterminant have an infinite number of answers or limits?
when i do the limit of (1 + x)^1/x as x approaches 0, this represents the indeterminant form of 1^infinity.
if you solve using hospitals rule, you get ln y = 1
which means y = e
however, if i plug in 2/x as the exponent, it still represents 1^infinity, but i get a different answer this time, which happens to be e^2.
and if i plugged in 3/x i would expect my answer to be e^3, so does it have an infinite number of limits?

The three problems you mentioned ARE three different problems that all lead to the same indeterminate form. However, you have correctly found the three different answers. As you can see, at first you don't know what the answer is even though it sort of seems like it should be the number one. That is why we call the form indeterminate. It MIGHT be one or it might be something else, totally not expected even ! We don't know what the answer is without investigating further.

If you'll rearrange the different problems you'll see that you have something to the first power, something to the second power and so on. That's why the answers are e^1, e^2, etc. The inner function has a limit of e and all the rest are that function raised to whatever power you are using.

Why do you think that (1 + x)^(1/x) is the same as (1 + x)^(2/x) an etc?