# E and the wrong thought process for limits

1. Jul 4, 2013

### Jbreezy

1. The problem statement, all variables and given/known data
So I keep evaluating these limits wrong. Something is wrong in my thought process.

2. Relevant equations
http://www.calcchat.com/book/Calculus-ETF-5e/ chapter 9, section 1, question 69
Consider
A_n = (1+k/n)^n

If I wanted to see if it converged or not I would just imagine what n did as it got really large. So k/n goes to 0. So you have limit n-->infty(1+ k/0)^n = 1? I know this is wrong. But when I see things like this I want to do something like that.
Why can't I?
Also the book makes u = k/n and in effect end up rewriting it as limit as u-->0 ( (1+ u)^1/u)^k = e^k.

I guess I don't really grasp the concept here. I don't understand how and why you know to rewriting this. I initially just said oh that is e but what was wrong. Can someone explain this.

3. The attempt at a solution

2. Jul 4, 2013

### jbunniii

The $n$ in the denominator of $k/n$, and the $n$ in the exponent, are pulling in opposite directions: the denominator is trying to make the terms smaller, and the exponent is trying to make them larger. Remarkably, as $n$ grows without bound, these two "forces" balance each other and the sequence converges to a positive number greater than 1 and less than infinity.

Your attempt was to allow the denominator of $k/n$ to grow while holding the exponent fixed. But that's not how the sequence is defined.

It is a fundamental fact that
$$e^x = \lim_{n \rightarrow \infty} \left(1 + \frac{x}{n} \right)^n$$
How you prove this depends upon your definition of $e^x$. One definition that is often used is
$$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$
With some effort, it's possible to show that these two limits are equal. See for example Baby Rudin, chapter 3, for the $x = 1$ case.

3. Jul 4, 2013

### jbunniii

To elaborate a bit further, a standard approach (again, following Rudin), is as follows:

(1) Prove that $\sum_{n=0}^{\infty} \frac{1}{n!}$ converges. Call the limit $e$. This sum also gives a straightforward way to estimate the value of $e$.

(2) Prove that $e = \lim_{n \rightarrow \infty} (1 + 1/n)^n$. (This is substantially harder!)

(3) Given (2), show that $e^x = \lim_{n \rightarrow \infty} (1 + x/n)^n$. This is pretty easy via the substitution indicated at your link.

An alternate approach is to define $e$ using the limit in (2), and prove (1) from (2).

How is $e$ defined in your course?

4. Jul 4, 2013

### Jbreezy

Hey thank you for your reply. $e$ is not defined in my course. I'm not in a formal course I'm working ahead in the summer. I just came across this one question. The question was given in the form $$A_n = \lim_{n \rightarrow \infty} \left(1 + \frac{k}{n} \right)^n$$
Then they do the substitution with u. I guess I'm having a hard time formulating my question. I want to know how you know when you see the equation I just wrote to say oh I need to substitute $$u= \frac{k} {n}$$

Because both are $$e$$ right? The form after the substitution and before. I guess my brain is missing the reason how I would know to do this. I don't get it. I feel like I don't know the reason. And who the heck is Rudin?

5. Jul 4, 2013

6. Jul 4, 2013

### Integral

Staff Emeritus
Things like this fall into the class of "Any 5 yr old can do it.... With 10 yrs experience.

7. Jul 4, 2013

### Jbreezy

I'm not sure if you are making fun of me or not.What does this mean?

8. Jul 4, 2013

### jbunniii

I assume that the author of the problem expects you to recognize the fundamental limit
$$\lim_{n \rightarrow \infty} \left(1 + \frac{1}{n}\right)^n = e$$
either by definition or as a nontrivial theorem, if $e$ was defined some other way. This is not a fact that you would casually derive in a routine homework problem.

If we assume that fundamental limit is known, then it isn't too hard to find the limit when we replace the $1$ in the numerator with $x$:
$$\lim_{n \rightarrow \infty} \left(1 + \frac{x}{n}\right)^n$$
Indeed, the solution shows how to do this, but I'll write it out in more detail in case that helps. Assuming $x > 0$, we can let $u = n/x$, and so $n = xu$. Also, $n \rightarrow \infty$ if and only if $u \rightarrow \infty$. Therefore, we can rewrite the above limit as
$$\lim_{u \rightarrow \infty} \left(1 + \frac{1}{u} \right)^{xu}$$
Since $(1 + 1/u)^{xu} = ((1 + 1/u)^u)^x$, we can simplify the above to
$$\left(\lim_{u \rightarrow \infty} \left(1 + \frac{1}{u} \right)^{u}\right)^x$$
But the limit inside the outermost set of parentheses is simply $e$, so we end up with $e^x$.

You can make a similar argument if $x < 0$, except in that case, $n \rightarrow \infty$ if and only if $u \rightarrow -\infty$.

Finally, if $x = 0$, then the limit is simply $1$, which is consistent with $e^0 = 1$.

He is the author of a commonly used real analysis book. If you continue studying math, you will probably encounter this book sooner or later, but no need to worry about it for now.

9. Jul 4, 2013

### jbunniii

No, I'm sure he is not making fun of you. What he means is, that the problem is quite simple if you are already aware of the very important and nontrivial fact that
$$\lim_{n \rightarrow \infty}\left(1 + \frac{1}{n}\right)^n = e$$
On the other hand, if you don't know this fact, the problem is practically impossible.

10. Jul 4, 2013

### Jbreezy

When I first saw the problem I said "Oh it is e" because I did know this.
$$\lim_{n \rightarrow \infty}\left(1 + \frac{1}{n}\right)^n = e$$

But because $$\frac{1} {n}\$$ was $$\frac{k} {n}\$$ it completely threw me off when I checked my solution and saw that is was $$e^k$$

I could not understand how I would know that I should substitute $u = k/n$

Thanks for the reply. Why do they call him Baby Rubin?

11. Jul 4, 2013

### jbunniii

LOL, it's the book, not the author, that is called Baby Rudin. That name refers to his undergraduate book, Principles of Mathematical Analysis. He also has a graduate book, Real and Complex Analysis, which is often called "Big Rudin" or "Adult Rudin." This isn't some nickname I made up - believe it or not, these terms are actually used quite commonly in math departments for these two books.

12. Jul 4, 2013

### Jbreezy

Haha. That's pretty funny. I was thinking how that is the weirdest thing for an academic to have a nickname like that.

Good books no?

13. Jul 4, 2013

### jbunniii

The goal was to transform the limit we are trying to find,
$$\lim_{n \rightarrow \infty}\left(1 + \frac{x}{n}\right)^n$$
in terms of a limit we already know,
$$\lim_{n \rightarrow \infty}\left(1 + \frac{1}{n}\right)^n$$
So by choosing $u = n/x$, we are able to transform $x/n$ into $1/u$.
They're good reference books, because they are very terse and contain good, clean proofs of the main results of calculus/analysis. However, the terseness means that they are also not a very user-friendly source for learning the material the first time around, unless they are supplemented by a good lecturer.

14. Jul 4, 2013

### Jbreezy

Ah I see. Like Apostol?

15. Jul 4, 2013

### jbunniii

Same level of rigor as Apostol, but much more terse. Imagine compressing most of volume 1 of Apostol's Calculus into about 200 pages, and adding some extra material. Also, delete every picture/plot/diagram, every easy exercise, and almost every example.

16. Jul 4, 2013

### Jbreezy

Ah. I see sounds lovely. Such a deceiving name "Baby Rubin."

What did you use it for?

17. Jul 4, 2013

### jbunniii

I use it as a reference. Fortunately, when I learned undergraduate analysis, my professor chose a different text, Robert Bartle's The Elements of Real Analysis. I am appalled to see that it is currently out of print, not least because the binding of my copy is totally trashed. It's a lovely book, at the same level of sophistication as Baby Rudin but much easier to learn from. However, it's also not as easy to find things quickly in Bartle's book, so I think Baby Rudin is a better reference.

18. Jul 4, 2013

### Jbreezy

Got ya. Well thanks for the help be easy.