# Limit at infinity that may or may not be e

1. May 13, 2010

### Marylander

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

Compute lim as x goes to infinity of (1+1/x^2)^x

2. Relevant equations

I know that lim at infinity (1+1/x)^x=e

I do not know if that is still valid with the x^2 there. I don't really think it is, but it's throwing me off.

3. The attempt at a solution

Beyond the above bit of knowledge, 1/x^2 goes to 0, leaving me with 1 to a power. So the limit should be 1, I believe, but I want to make absolutely sure.

2. May 13, 2010

### Staff: Mentor

$$[1^{\infty}]$$ is one of several indeterminate forms. The way to evaluate lim (1 + 1/x)^x is by letting y = (1 + 1/x)^x, and then taking the natural log of both sides and getting to something that you can use L'Hopital's Rule on. Do the same thing with your limit.

3. May 13, 2010

### Marylander

1^infinity is indeterminate? How? 1 is always 1, isn't it?

EDIT: And sorry, the limit I'm trying to do isn't (1+1/x)^x, it's (1+1/x^2)^x. Is it the same process?

EDIT2: Okay.... so lim (1+1/x^2)^x is also e? Or at least that's what I got, anyway.

Last edited: May 13, 2010
4. May 13, 2010

### Staff: Mentor

Yes.
If you mean Why? the reason is that a limit expression in which the base is approaching 1 but the exponent is becoming infinitely large can turn out to be anything.
Well 1 is always 1, that's true, and 1n = 1 provided that n is a finite number.
Yes.

5. May 13, 2010

### Marylander

Very impressive monosyllables there.

And thanks for the help. I see how to deal with this. Which is good, because it's just about inevitable that I'll have to for my final.

If this board let me rep you or something I would, but you're just have to make do with the much-sought-after textual thanks.

6. May 13, 2010

### Dick

Be careful. The process for finding lim (1+1/x^2)^x is the same as (1+1/x)^x, but the answer isn't the same. It's not 'e'. Can you show how you got that?

7. May 13, 2010

### Marylander

Great...

Sure:

elim(x(ln(1+1/x2))

lim xln(1+1/x2)=lim ln(1+1/x2)/(1/x)

L'Hopital:

Lim 1/(1+1/x2)*(-x-2)/-x-2

Lim 1/(1+1/2)=1

e1

EDIT: Ah. There's where I messed up. It should be...

Edit2: elim 2/(x(1+1/x2)

That limit is 0, so it's e0=1, which was my original answer.

Last edited: May 13, 2010
8. May 13, 2010

### Dick

That's the right idea, but the derivative of 1/x^2 is -2/x^3. Not -x^(-2).

9. May 13, 2010

### Marylander

Yeah, I caught that after I submitted it. Absent-minded simple messups.

The edits have my corrected answer. The limit is 1, correct?

10. May 13, 2010

### Dick

Correct.

11. May 13, 2010

### Marylander

Thanks again for all the help.