# Really dumb question but

1. May 9, 2012

### tahayassen

1. The problem statement, all variables and given/known data

Calculate the value of e.

2. Relevant equations

3. The attempt at a solution

If I substitute n=infinity into that limit, wouldn't I get e = 1?

(1 + 1/(infinity))^infinity

As the denominator of 1/infinity approaches very large values, 1/infinity approaches zero.

Therefore, (1 + 0)^infinity

e = 1

2. May 9, 2012

### SammyS

Staff Emeritus
No, you won't get 1 .

Try some successively larger values for n to see what happens !

3. May 9, 2012

### tahayassen

Ya, but I don't understand why.

4. May 9, 2012

### sharks

Not so long ago, i did the same mistake as you did in post #1.

Basically, you're trying to find the limit of: $\lim_{n\to \infty}n^n$
$$Let\; y=\left( 1+ \frac{1}{n} \right)^n$$ Hint: Take logs on both sides.

5. May 9, 2012

### Steely Dan

Try writing out the polynomial for a given $n$:
$$s_n = \left(\frac{1}{n}\right)^n (1+n)...(1+n) = \left(\frac{1}{n}\right)^n (1 + a_1 n + a_2 n^2 + ... + a_{n-1} n^{n-1} + a_n n^n)$$
where the coefficients are simply those of Pascal's triangle. Obviously as $n\rightarrow \infty$, the former terms are small compared to the latter, so
$$s_n \approx \frac{1}{n^n} ((n-1) n^{n-1} + n^n),$$
since the second to last term in a row of Pascal's triangle is always the number of the row minus one. Therefore
$$s_n \approx \frac{1}{n^n} (2 n^n - n^{n-1}) = 2 - \frac{1}{n}$$
Obviously the latter term drops off as $n$ gets large, but the contribution it gave to the term of order $n^0$ sticks behind. Try it with keeping the contribution to that order with three terms, four, etc.

Edit: And I don't think this is a dumb question. I had to come up with that reasoning when I just read your post. I suppose it serves to illuminate that sometimes our intuition about the behavior of large numbers can lead us astray!

Last edited: May 9, 2012
6. May 9, 2012

### tahayassen

I took the log of both sides, and then I moved the n exponent to the coefficient, so I got:

logy = nlog(1+1/n)

But I'm not sure where to proceed from here.

Wow. I'm totally lost. I didn't even understand the very first step you did.

If I understand correctly, Pascal's triangle is:

So where did you get $$s_n = \left(\frac{1}{n}\right)^n (1+n)...(1+n)$$ from?

7. May 9, 2012

### Steely Dan

The first step I did is to pull out $n^{-n}$ from the polynomial, leaving $s_n = n^{-n} (1+n)^n$. Pascal's triangle gives you the coefficients of that latter polynomial, whose lowest order term is surely 1 and whose highest order term is surely $n^n$.

8. May 9, 2012

### Staff: Mentor

It's not legal to substitute ∞ into an expression and do arithmetic on it.
This is one of several indeterminate forms, which include [0/0], [±∞/∞], [∞ - ∞], and [1]. These show up in the context of limits, and indicate that some more work needs to be done.

9. May 9, 2012

### tahayassen

y=(1+n^-1)n
=n^-n(1+n)^n

How did you go from the first step to the second step?

What if I do:

Let m = 1/n

lim as m approaches 0 (1+m)^(1/m)
= (1+0)^(1/0)
= 1^(1/0)

10. May 9, 2012

### Staff: Mentor

It's not legal to divide by zero, either.

11. May 9, 2012

### tahayassen

So how would you solve this limit without using the approximation method then?

12. May 9, 2012

### Staff: Mentor

Continue with what you were doing earlier...

log(y) should be ln(y).
Rewrite the right side as [ln (1 + 1/n)]/(1/n).
Take the limit of both sides. If necessary, use L'Hopital's Rule.

13. May 9, 2012

### tahayassen

I give up.

I'm trying to learn L'Hopital's Rule. Learning Pascal's Triangle was already hard enough. :(

edit: I'll keep trying, but this is going to take me some time.

14. May 9, 2012

### sharks

L'Hopital's Rule isn't too complicated: here is the formal definition
Basically, it relates to fractions and if you put the value of the limit in the numerator and denominator, and if you get 0/0 or ∞/∞, (which are called indeterminate forms) then you will need to use L'Hopital's Rule. The rule simply requires that you differentiate both the numerator and denominator, independently. In some cases, you'll need to repeat the procedure.

15. May 9, 2012

### tahayassen

I just noticed at the end of this: not only did I divide by zero, but I still get 1^infinity at the end, which is an indeterminate form. Opps!

Back on track, okay, so I did that rule, but I still ended up with a value of 1! D=

http://i2.lulzimg.com/4cb849a256.png [Broken]

$$y\quad =\quad { (1+{ n }^{ -1 }) }^{ n }\\ lny\quad =\quad { ln(1+{ n }^{ -1 }) }^{ n }\\ lny\quad =\quad nln(1+{ n }^{ -1 })\\ \underset { n\rightarrow \infty }{ lim } lny\quad =\quad \underset { n\rightarrow \infty }{ lim } \frac { ln(1+{ n }^{ -1 }) }{ { n }^{ -1 } } \\ \underset { n\rightarrow \infty }{ lim } lny\quad =\quad \underset { n\rightarrow \infty }{ lim } \frac { \frac { 1 }{ 1+{ n }^{ -1 } } ({ -n }^{ -2 }) }{ { -n }^{ -2 } } \\ \underset { n\rightarrow \infty }{ lim } lny\quad =\quad \underset { n\rightarrow \infty }{ lim } \frac { 1 }{ 1+{ n }^{ -1 } } \\ \underset { n\rightarrow \infty }{ lim } lny\quad =\quad \frac { 1 }{ 1+{ \infty }^{ -1 } } \\ \underset { n\rightarrow \infty }{ lim } lny\quad =\quad \frac { 1 }{ 1+0 }$$

Last edited by a moderator: May 6, 2017
16. May 9, 2012

### sharks

So, you've reached: $$\lim_{n\to \infty} \ln y = 1$$
But, $$\lim_{n\to \infty} \ln y = \ln \lim_{n\to \infty} y$$
Now, can you find $\lim_{n\to \infty} y$?

17. May 9, 2012

### Staff: Mentor

OK, since lim ln(y) = 1, then y = ?

18. May 9, 2012

### tahayassen

ln * lim(y) = 1
lim(y) = e^1
= e

But I still haven't found a value. :|

19. May 9, 2012

### sharks

Here, e, represents the exponential value. Use your calculator. What is the value of $e^1$?

Last edited: May 9, 2012
20. May 9, 2012

### tahayassen

The original question was:

Calculate the value of e.

Essentially, what we did was:

e = e
= 2... (using the calculator function)

But how did the calculator get that value?