$ \lim_{x \rightarrow \infty} e^{1/x} = a $ is not equivalent with $ \lim_{x \rightarrow \infty} a^x = e$? I have problems understanding, why
$$ \lim_{x \rightarrow \infty} e^{1/x} = a  $$ is not equivalent with
$$ \lim_{x \rightarrow \infty} a^x = e.   $$
In the first case there is a solution $a=1$, and in the second it does not work.
 A: You're misunderstanding the notation. The $\lim$ notation works on expressions, not entire equations. So
$$ \lim_{x\to\infty} e^{1/x} = a \qquad\text{means}\qquad
\left(\lim_{x\to\infty} e^{1/x} \right) = a $$
And
$$ \lim_{x\to\infty} e = a^x \qquad\text{would mean}\qquad
\left(\lim_{x\to\infty} e \right) = a^x $$
but that is nonsense because $x$ does not exist at all outside the limit, and in particular not to the right of the equals sign.
This goes deeper than just now to notate things, because once you realize that "$\lim_{x\to\infty} e^{1/x}=a$" does not contain (or have anything to do with) the equation "$e^{1/x}=a$", you should be cured of trying to replace the latter with the equivalent equation $e=a^x$. The replacement doesn't make sense because none of the things you try to replace for each other are actually there.
A: I think your basic problem comes from the standard equations $$e^{1/x} = a \Leftrightarrow a^{x} = e$$ which is perfectly valid. Note that from the above it follows that $a$ depends on $x$.
When you take limit of $e^{1/x}$ as $x \to \infty$ the result is not something which depends on $x$, rather it is independent of $x$. You have understand this very clearly. When you take limit you get something which is independent of the limit variable. In this case the limit of $e^{1/x}$ is $1$ as $x \to \infty$. So there is no variable $a$ (dependent on $x$) once you use the limit operation. So it does not make sense to consider the $a$ of $a^{x}$ in the same manner. The equation $a^{x} = e$ requires that $a$ be dependent on $x$, but when you consider the limit of $a^{x}$ when $x \to \infty$ you have to deal with $a$ as independent of $x$.
BTW in contrast here is a more interesting and very surprising result when we deal with two variables. Let $n$ be a positive integer and $x$ be a real number. Then we have the following equation $$\left(1 + \frac{x}{n}\right)^{n} = y\,\Leftrightarrow\, x = n(y^{1/n} - 1)\tag{1}$$ and this relations holds even if we take limits as $n \to \infty$ and we have $$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} = y\, \Leftrightarrow\, x = \lim_{n \to \infty}n(y^{1/n} - 1)\tag{2}$$ Note that there is no general rule for going from one equation like $(1)$ to another equation like $(2)$ but there are certain interesting examples like the above where such inferences are valid.
A: They are not equivalent simply because the in first limit you wrote is $a=1$, in the second the limit is


*

*$\infty$ if $|a|>1$ 

*$0$ if $|a|<1$

*$1$ if $a=1$

*doesn't exist if $a=-1$

A: Those are not equivalent.
For $f(x)$ a real function, the limit of $f$ as $x\to\infty$ being $a$ means that, for all $\varepsilon>0$, there exists $\bar x\in\mathbb{R}$ such that $|f(x)-a|<\varepsilon$ whenever $x>\bar x$. That is:
$$\lim_{x \to \infty} f(x) = a \Leftrightarrow \forall \varepsilon > 0, \ \exists \bar x\in\mathbb{R} \ \mbox{ such that }\; \forall x > \bar x, \ |f(x) - a| < \varepsilon.$$
This means that you can get as close as you want to $a$. But it does not imply that you will achieve it for a particular value of $x$. In other words, there is no guarantee that there exists $x\in\mathbb{R}$ such that $f(x)=a$.
In your case, your first limit means that the bigger $x$ is, the closer $e^{1/x}$ gets to $a=1$. But it never gets exactly to $1$. 
On the other hand, in your second limit you are given $a^x = 1^x =1$. You are not considering where does that $1$ come from. It could be the limit of $f(x)=e^{1/x}$, $f(x)=\frac{1}{x}+1$, or even simply $f(x)=1$! It does not matter; you just have exactly $1$. Thus, $1^x=1$ for all $x$ and the limit as $x\to\infty$ is also equal to $1$.
