Limit at infinity that may or may not be e

Marylander
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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.
 
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Marylander said:

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.
 
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:
Marylander said:
1^infinity is indeterminate?
Yes.
Marylander said:
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.
Marylander said:
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.
Marylander said:
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.
 
Very impressive monosyllables there. :smile:

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.
 
Marylander said:
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?
 
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:
Marylander said:
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

That's the right idea, but the derivative of 1/x^2 is -2/x^3. Not -x^(-2).
 
Yeah, I caught that after I submitted it. Absent-minded simple messups.

The edits have my corrected answer. The limit is 1, correct?
 
  • #10
Marylander said:
Yeah, I caught that after I submitted it. Absent-minded simple messups.

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

Correct.
 
  • #11
Thanks again for all the help.
 
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