# If $n = xm$ and $n \rightarrow \infty$, then $m \rightarrow \infty$?

Did Stewart prove the titular result, also underlined in green in the image below, either by himself or as an exercise, in *Calculus Early Transcendentals* or in the normal version *Calculus*?

I attempted the proof, but I got nonplussed. $\infty = \lim\limits_{n \rightarrow \infty} n = \lim\limits_{n \rightarrow \infty} xm $. Now what? The limit is for $n$, but my variables are $x$ and $m$.

- Show that $\lim\limits_{n \rightarrow \infty} (1 + \dfrac xn)^n = e^x$ for any $x > 0$.

Transcription of image follows below:

56.Let $m = n/x$. Then $n = xm$, and as $n \to \infty$, $m \to \infty$.Therefore, $\displaystyle \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = \lim_{m \to \infty} \left( 1 + \frac{1}{m} \right)^{mx} = \left[ \lim_{m \to \infty} \left( 1 + \frac{1}{m} \right)^m \right]^x = e^x$ by Equation 6.

^{James Stewart, Calculus Early Transcendentals, 7 ed. (2011) p. 223.}

By the way, if I was writing the solution, I would've started with the game plan: rewriting $(1 + x/n)$ in the form $(1 + 1/\mathrm{something})^{\mathrm{something}}$ in order to apply $e = \lim\limits_{something \rightarrow \infty} (1 + 1/\mathrm{something})^{\mathrm{something}}$. It's superfluous to change variable to $m$. Just rewrite $(1 + x/n)$ as $\left(1 + \dfrac{1}{\frac nx}\right).$

But if you pined to change variable, I would've commenced with defining $\dfrac xn$ as $\dfrac 1 m$ which is more intuitive than "let $m = n/x$", because only $\dfrac xn$ shows up explicitly in the question. The question doesn't manifest $n/x$.

## 1 answer

[…] I would have commenced with defining $\dfrac{x}{n}$ as $\dfrac{1}{m}$, which is more intuitive than "let $m = n/x$".

I confess I don't see the difference between the two statements. In changing the variable in the limit from $n$ to $m$, you keep $x$ fixed; after all, you are trying to prove that for each fixed $x > 0$, the following limit is true: $$ \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = e^x. $$

The same is true when you "let $m = n/x$". It is not correct to say that in this case "my variables are $xm$" (I think you meant to say "$x$ and $m$", or "variable is"?) because $x$ has been fixed right at the beginning as an arbitrary positive real number.

Admittedly, Stewart expects this to be clear from the phrasing of the exercise itself, and does not begin the solution with, "Let $x > 0$ be fixed". At least, it is not unusual for a solutions manual to be terse on such points, in my experience.

I disagree that the change of variable is superfluous. How will you proceed after you rewrite $x/n$ as $1/(n/x)$? You will probably multiply and divide by $x$ in the exponent to get $$ \lim_{n \to \infty} \left( 1 + \frac{1}{n/x} \right)^{\frac{n}{x} \cdot x} = \left[\lim_{n \to \infty} \left( 1 + \frac{1}{n/x} \right)^{n/x} \right]^x. $$ This is of the form you want: $(1 + 1/\mathrm{something})^{\mathrm{something}}$ and the limit as that "$\mathrm{something}$" goes to infinity. Except, that's not what the limit looks like! You want the limit as $n/x$ goes to infinity, not as $n$ goes to infinity. But $n/x$ goes to infinity if and only if $n$ goes to infinity (when $x > 0$ is fixed) so we can change the limit to $n/x \to \infty$ from $n \to \infty$. But, this is just the change of variable process without it being spelt out. So, it's actually not at all superfluous.

Adding in response to the updated version of the question and the comments below:

The statement that you wish to prove is along the lines "*something* is true for all $x$". Or, no matter what $x$ you pick, *something* will hold for that $x$. So, a proof of the statement must begin by letting $x$ denote an arbitrary element and then showing that *something* is indeed true for that arbitrary $x$. In other words, $x$ is fixed (and arbitrary) during the proof.

You already recognise this, at least intuitively, when you observe that as $n \to \infty$ we have $n/x \to \infty$ too. This is made explicit when writing down the change of variable from $n$ to $m$: the varying parameter just switches from being $n$ to being $m$, while $x$ stays fixed (but arbitrary) throughout.

Regarding the objection to the introduction of the change of variable itself (or, at least the way it is introduced), I feel it is more on pedagogical grounds. And, opinions will vary here. It is good that you recognize that you want to get an expression of the form $1 + 1/\mathrm{something}$, and so it makes sense that you would like to define $x/n = 1/m$. On the other hand, a student who does not recognise this might find this a lot more confusing: why are the original and new variables both in the denominator? After all, this is a completely nonstandard way to write down the linear equation $m = n/x$.

This is just one argument to show that there is no single method that works for everyone. So, I can't find fault with Stewart in his presentation of the change of variable, nor yours. But it would be good to keep an open mind about the benefits of different ways of presenting the same material.

#### 5 comments

Thanks. (1) "I confess I don't see the difference between the two statements." You're correct. There isn't. I edited my post to elaborate. (2) "my variables are " (I think you meant to say "x and m", or "variable is"?) Yes. Thanks. I corrected this.

(3) "x has been fixed right at the beginning as an arbitrary positive real number." You spotted my issue! Can you please elaborate? Why can you just assume or deudce that x is fixed? As you wrote, "Stewart expects this to be clear from the phrasing of the exercise itself, and does not begin the solution with, "Let x > 0 be fixed"". This isn't clear to me. IMO, Stewart ought to have written this!

Can you please respond in, by editing, your answer? Comment chains are cumbersome to read.

@Technologicallyilliterate Sure, I'll try to add something tomorrow.

@Technologicallyilliterate Added a few words to my answer.

## 5 comments

Do you have a definition of $\lim_{a \to \infty} f(a) = \infty$ in first order logic (i.e. as a simple statement with $\exists$ and $\forall$)? — Peter Taylor 4 months ago

@PeterTaylor Yes. https://i.imgur.com/mA64hx0.jpg — VLDR 4 months ago

That won't do; that definition requires the limit to be a finite number, but you're working with an infinite limit. Also, downvoted because once again, you're using images of text instead of writing your question such that no images are required. — r~~ 4 months ago

@r~~ Do you mean https://i.imgur.com/i5jxtFW.jpg? "you're using images of text instead of writing your question such that no images are required" Images ARE required. I don't want to mis type the solution manual. I want to show you exactly what it says. — VLDR 4 months ago

Just transcribe the content of the image in addition to posting the image. It's good that you want to ensure that you're not mistyping the solution manual, but presenting text as images makes it difficult or impossible for people with accessibility issues to know what you're referring to. — Bahudari Ragam 4 months ago