Limit at infinity that may or may not be e

Click For Summary

Homework Help Overview

The problem involves computing the limit as x approaches infinity of the expression (1 + 1/x^2)^x. Participants are exploring whether the known limit of (1 + 1/x)^x, which approaches e, applies in this case with the x^2 in the denominator.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the behavior of the expression as x approaches infinity, questioning the validity of applying the known limit of (1 + 1/x)^x to the modified expression. There is an exploration of the indeterminate form 1^infinity and its implications.

Discussion Status

The discussion has seen various attempts to clarify the limit's behavior, with some participants suggesting that the limit might still be e, while others caution that the answer could differ. There is a recognition of the need to apply L'Hopital's Rule and logarithmic transformations to evaluate the limit correctly.

Contextual Notes

Participants note the potential confusion arising from the change in the denominator from x to x^2, leading to differing interpretations of the limit's outcome. There is also mention of the indeterminate nature of the expression, which requires careful analysis.

Marylander
Messages
8
Reaction score
0

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.
 
Physics news on Phys.org
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.
[tex][1^{\infty}][/tex] 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.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
3
Views
1K
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K