Limit at infinity that may or may not be e

[Marylander]

Homework Statement

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

Homework 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.

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.

 

[Mark44]

Homework Statement

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

Homework 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.

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.

$$[1^{\infty}]$$ 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.

 

[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.

 

[Mark44]

1^infinity is indeterminate?

Yes.

How?

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.

1 is always 1, isn't it?

Well 1 is always 1, that's true, and 1ⁿ = 1 provided that n is a finite number.

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?

Yes.

 

[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.

 

[Dick]

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.

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?

 

[Marylander]

Great...

Sure:

e^{lim(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: e^{lim 2/(x(1+1/x2)}

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

 

[Dick]

Great...

Sure:

e^{lim(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

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

 

[Marylander]

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

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

 

[Dick]

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

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

Correct.

 

[Marylander]

Thanks again for all the help.