Generalization of $e$'s limit form

Question

We know that

$$e = \lim\limits_{n \to \infty} \Bigl( 1 + \frac{1}{n}\Bigr)^n \tag{1}$$

I don't know how to evaluate that limit to get that result, but assuming that I know the above result, can I conclude the following?

$$x = \lim\limits_{n \to \infty} \Bigl( 1 + \frac{1}{an}\Bigr)^n = \lim\limits_{n \to \infty} \Bigl( 1 + \frac{1}{n}\Bigr)^\frac{n}{a} = e^\frac{1}{a}\ where\ a \ne 0$$

EDIT: Mees has pointed out in the comments that this result is correct. But can someone provide a proof / justification? Using $(1)$, how could I arrive at these equalities without having to solve the limit itself?

Answer 1

Note that

\begin{eqnarray} e^x &=& \left[\lim_{n\to\infty} \left(1 + \frac{1}{n} \right)^n \right]^x &=& \lim_{n\to\infty} \left(1 + \frac{1}{n} \right)^{nx} \\ &\stackrel{y = nx}{=}& \lim_{y\to\infty} \left(1 + \frac{x}{y} \right)^y \\ \end{eqnarray}

Answer 2

This answer assumes that $n$ is a positive integer variable.

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If $a$ is a non-zero rational number then the result you seek can be obtained from the assumption $(1+(1/n))^{n}\to e$ via simple algebraic manipulation.

When $a$ is irrational then the result can not be obtained via algebraic manipulation, but rather one has to take into account some definition of $x^{y}$ for irrational $y$. Then one can show that the limit exists and is equal to $e^{1/a}$. Also I find it bit strange that you want to work with $a$ in denominator rather than directly handling this by putting $b=1/a$ and using $b$ in numerator (this one is more common).