Limits Involving $e$ - Is this acceptable?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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

$$\lim _{(x,y) \to (x_0,y_0)}f(x,y) = \lim_n \lim _m f(x_n,y_m) = \lim_m \lim _n f(x_n,y_m) = f(x_0,y_0)$$

Answer 4

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 $$ \lim_{n\to\infty}\left(1+\frac1n\right)^n=1\tag3 $$