Calculate the limit

[LCSphysicist]

Homework Statement:

.

Relevant Equations:

(1 + a^x)^(a^(-x))

(1 + a^x)^(a^(-x))
Let's assume a, say, two.

the limit of it, with x tending to infinity,

is one, but i was thinking...
Calling 2^x by a, we have that when x tend to infinity, so do a, So:

that is euler number no? Contradictory... where am i wrong?

 

[Eclair_de_XII]

one

That's what I got. The way I usually deal with limits of this form is to express it as a function of the exponential function and the natural logarithm, then use l'Hopital's rule on the power.

 

[LCSphysicist]

That's what I got. The way I usually deal with limits of this form is to express it as a function of the exponential function and the natural logarithm, then use l'Hopital's rule on the power.

Yes but, what is wrong in the second case?

 

[Eclair_de_XII]

I think that the problem is that $a$ is not bounded, and it would not be the same as the limit for Euler's number if you tried to put it that way. For example, if you let $a=\frac{1}{n^2}$, then as $a\rightarrow \infty$, $n^2\rightarrow 0^+$. Then your second limit becomes:

$\lim_{a \rightarrow \infty}(1+a)^{\frac{1}{a}}=\lim_{n^2\rightarrow 0}(1+\frac{1}{n^2})^{n^2}\neq e$

Maybe if you switched around the negative signs in the equation in the Relevant Equations part of your post, it would be some function of Euler's number.

 

[Delta2]

Euler's number is $\lim (1+\frac{1}{n})^n$ as $n\to \infty$ or equivalently $\lim (1+n)^{\frac{1}{n}}$ as $n\to 0$.

BUT you have $\lim (1+n)^{\frac{1}{n}}$ as $n\to \infty$

 

[LCSphysicist]

Euler's number is $\lim (1+\frac{1}{n})^n$ as $n\to \infty$ or equivalently $\lim (1+n)^{\frac{1}{n}}$ as $n\to 0$.

BUT you have $\lim (1+n)^{\frac{1}{n}}$ as $n\to \infty$

Oh so i was wrong to think that the third therm was too equivalent.
I really messed me up, the terms between the parentheses need to tend to one, not to infinity... thank you both guys