Limits Involving $e$ - Is this acceptable? I'm currently studying a course in real analysis and have come across the following well known limit: $$\lim _{n\to\infty}\left(1 + \frac{1}{n}\right)^n=e$$
My question is: If you were given a question that asked you to compute $\lim _{n\to\infty}e^{1/n}$, would it be acceptable to do the following? $$\lim _{n\to\infty}e^{1/n}=\lim _{n\to\infty}\Bigg(\left(1 + \frac{1}{n}\right)^n\Bigg)^{1/n}=1$$
I know you could just take the limit of the power of $e$ in this case, but I'm just thinking for harder questions that I have found.
 A: No.  You are trying to sneakily replace a nested limit with a diagonal limit.
You have $\lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^n = \mathrm{e}$, so anywhere you have "$\mathrm{e}$", you can replace it with "$\lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^n$" (possibly changing the limit variable to avoid using an already bound variable).
In $\lim_{n \rightarrow \infty} \mathrm{e}^{1/n}$, $n$ is already bound, so we have to use a different placeholder variable in our substitution.  Let's use $m$.  We get
$$  \lim_{n \rightarrow \infty} \left( \lim_{m \rightarrow \infty} \left( 1 + \frac{1}{m} \right)^m \right)^{1/n}  $$
where I have wrapped our substitution in parentheses to avoid any possible ambiguity in meaning from the 2-dimensional notation.
Notice the process that this describes:


*

*For each member in a sequence of $n$s which goes to infinity, calculate the $1/n^{\text{th}}$ power of 


*

*For each member in a sequence of $m$s (which may be a different sequence for each of the values of $n$ we use) which goes to infinity, 


*

*Evaluate $\left( 1 + \frac{1}{m} \right)^m$


*Finding its limit as $m \rightarrow \infty$.


*Finding its limit as $n \rightarrow \infty$.


This is not the same process as the single limit you write.  In general, the two variable limit can exhibit a range of behaviours that the diagonal limit (the limit along the diagonal line $m = n$) does not.
A: It seems to me that the "more general" question you are asking is:

If $a_n, b_n$ are sequences, and $$\lim_{n \to \infty} a_n = L$$ is it
  true that $$\lim_{n \to \infty} L^{b_n} = \lim_{n \to \infty}
 a_n^{b_n}.$$

I think this is a good question! I remember early in my career having similar thoughts, and not knowing how to decide if I could or not. But, this questions is still a math question so we should do what we always do, either prove that you can or exhibit a counter example which proves you cannot. So here we go.
In general, no you cannot. There might be special times where you can, but it would require justification.
As a counter example take $a_n = n^{1/n}$. Then (it is known and easy to prove) $$\lim_{n \to \infty} a_n = 1.$$ Let $b_n = n.$ Then we have $$\lim_{n \to \infty}\left[\lim_{j \to \infty} a_j \right]^{b_n} \lim_{n \to \infty} 1^{b_n}=\lim_{n \to \infty} 1^n = 1,$$ but $$\lim_{n \to \infty} a_n^{b_n} = \lim_{n \to \infty} (n^{1/n})^n = \lim_{n \to \infty} n = \infty.$$
In your original example however, you could (without proof) write it as $$\lim_{n \to \infty} \left[ \lim_{j \to \infty}(1+\frac{1}{j})^j \right]^{(\frac{1}{n})},$$ but in most cases, writing it in that way will not benefit the situation. But, it is correct.
A: Consider the limit
$$
\lim_{n\to\infty}\left(1+\frac1n\right)=1\tag1
$$
and the limit
$$
\lim_{n\to\infty}\color{#C00}{1}^n=1\tag2
$$
If we were to substitute $(1)$ into the place of $\color{#C00}{1}$ in $(2)$ and slide the limit outside as is done in the question, we would get
$$
\lim_{n\to\infty}\left(1+\frac1n\right)^n=1\tag3
$$
A: This type manipulation in general can be useful. In certain instances you can justify them by invoking continuity of some underlying function.
In this case one may consider the function $f(x)=x^y$ which is continuous on $(0,\infty)\times (-\infty,\infty)$.
Now, the different expressions you have are different paths of the limit of $f$ when $v\rightarrow (e,0)$, all of which must be equal to $f(e,0)$. 
In general if $f(x,y)$ is continuous at $(x_0,y_0)$ then for all sequences $x_n\rightarrow x_0$ and $y_m\to y_0$ we have 
$$\lim _{(x,y) \to (x_0,y_0)}f(x,y) = \lim_n \lim _m f(x_n,y_m) =  \lim_m \lim _n f(x_n,y_m) = f(x_0,y_0)$$
