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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?

Peeyush Kushwaha
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  • Yes, you can. $ $ – Mees de Vries Sep 17 '17 at 10:59
  • @MeesdeVries thanks for letting me know it's correct. I'm now editing my question to maybe ask for a proof or how I could arrive at those results – Peeyush Kushwaha Sep 17 '17 at 11:01
  • I think $a\neq0$. – Nosrati Sep 17 '17 at 11:03
  • @MyGlasses yes. Clarified it in the question – Peeyush Kushwaha Sep 17 '17 at 11:04
  • Hint: take the logarithm of the expression and use that $\log$ is differentiable at point $x=1$. – Gribouillis Sep 17 '17 at 11:05
  • "I don't know how to evaluate that limit to get that result" You don't have to. The limit exists (that must be proven, of course), and whatever it is, we give it the name $e$. – Arthur Sep 17 '17 at 11:21
  • @Arthur yes. But if I knew how to evaluate the original limit, I could easily evaluate the similar form. – Peeyush Kushwaha Sep 17 '17 at 11:28
  • Just to be specific, does $n$ in your question represent a positive integer variable (in which case question is non-trivial and interesting) or does it represent a real variable (in which case the question is trivial)? – Paramanand Singh Sep 17 '17 at 12:56
  • We can show that $\lim_{t\to \infty}(1+1/t)^t=e$ when $t$ is allowed to take any value in $\mathbb R^+.$.... For a>0 we have $(1+1/an)^n=[(1+1/t)^t]^{1/a}$ where $t=an.$ Since $t\to \infty$ as $n\to \infty$ and since the function $g(x)=x^{1/a}$ is continuous for $x>0$ and for fixed $a>0,$ therefore $\lim_{n\to \infty}(1+1/an)^n=$ $\lim_{t\to \infty}[(1+1/t)^t]^{1/a}=$ $[\lim_{t\to \infty}(1+1/t)^t]^{1/a}=$ $e^{1/a}.$ – DanielWainfleet Sep 17 '17 at 12:58
  • @DanielWainfleet: my point is that handling the limit when $n$ is integer is simpler from a conceptual point of view. If $n$ is real then it begs the development of theory of general power $x^{y} $ independent of $\log$ and $\exp$. This is possible but difficult. – Paramanand Singh Sep 17 '17 at 13:01
  • @ParamanandSingh I don't think it matters here, because in this question I wanted to manipulate the limits themselves and know how I can obtain $e^\frac{1}{a}$ from the known result and not dive into specifics of proving that result itself – Peeyush Kushwaha Sep 18 '17 at 16:46
  • The manipulation which you speak of (like the one in accepted answer) is valid only when $n$ is a real variable. Those do not apply when $n$ is integer. That's makes all the difference. BTW most common approaches do not prove equation $(1)$, rather take it as a definition with $n$ being a positive integer. – Paramanand Singh Sep 18 '17 at 18:54

2 Answers2

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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}

M. Winter
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caverac
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    Note: the map $x\mapsto x^n$ is continuous on $\Bbb R^+$. Therefore it commutes with the limit here. – M. Winter Sep 17 '17 at 11:44
  • The last expression, with the restriction $y=nx,$ is equal to $(1+1/n)^{nx} $ so I don't know how this is a solution. Am I missing something? – DanielWainfleet Sep 17 '17 at 12:08
  • @DanielWainfleet You can change the name of the variable $y$ back to$n$. It is just a label – caverac Sep 17 '17 at 12:15
  • It appears that you assume the limit $$\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^{x}=e$$ where $x$ is a real variable instead of the given limit in question which involves a positive integer variable $n$. The problem is trivial with the assumption you have made. – Paramanand Singh Sep 17 '17 at 12:29
  • @ParamanandSingh Maybe you'd benefit from reading the definition of $\lim_{x \to \infty}f(x)$. In particular de fact that $n$ is not necessarily an integer – caverac Sep 17 '17 at 12:51
  • By convention the $n$ in $n\to\infty$ is considered to be an integer. But if you assume the $n$ to be a real variable then the problem is trivial as your answer shows. I nowhere intended to mean that your answer is wrong. – Paramanand Singh Sep 17 '17 at 12:53
  • @ParamanandSingh intuitively speaking, it means "take a very large value of $n$". You can take such value as large as you want, doesn't have to be an integer – caverac Sep 17 '17 at 13:01
  • I did not want to be explicit (because your answers speak on your behalf here on MSE), but I do know the meaning of limit and I also know the difference between limit of functions of real variable and limit of functions of an integer variable (these are more properly called sequences) and I was just emphasizing this crucial difference which matters a lot for this question. I hope you get my point. – Paramanand Singh Sep 17 '17 at 13:05
  • @ParamanandSingh I most certainly get your point. I'm also afraid that any answer I give you at this point will not help you. I can delete my post if you think it just contributes to the confusion and give room to more skilled users to answer your questions – caverac Sep 17 '17 at 13:12
  • @ParmanandSingh it seems you're thinking my question deals with how to evaluate the limit $e = \lim\limits_{n \to \infty} \Bigl( 1 + \frac{1}{n}\Bigr)^n$, but I've mentioned in my question that the limit is is to be assumed and what I'm interested in is that if I know it, how can I arrive at the result: $x = \lim\limits_{n \to \infty} \Bigl( 1 + \frac{1}{an}\Bigr)^n = \lim\limits_{n \to \infty} \Bigl( 1 + \frac{1}{n}\Bigr)^\frac{n}{a} = e^\frac{1}{a}\ where\ a \ne 0$ - which I've guessed but it stands without a proof or deduction – Peeyush Kushwaha Sep 18 '17 at 16:51
  • @PeeyushKushwaha: I have nowhere talked about proving equation $(1)$ in my answer or in my comments. Equation $(1)$ of your question (with $n$ as a positive integer) is generally the starting point which can be taken as a definition of number $e$ and then an entire theory of exponential and logarithmic functions can be developed from this assumption. – Paramanand Singh Sep 18 '17 at 19:01
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This answer assumes that $n$ is a positive integer variable.

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).

Paramanand Singh
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