Mastering Limits: Solving \lim_{x\rightarrow 0+} x^{x^2} Using L'Hopital's Rule

  • Thread starter Thread starter laker88116
  • Start date Start date
  • Tags Tags
    Limit
Click For Summary

Homework Help Overview

The discussion revolves around evaluating the limit \(\lim_{x\rightarrow 0+} x^{x^2}\). Participants are exploring the application of L'Hôpital's Rule and the interpretation of the expression.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to start and mentions the need to rewrite the limit in a fraction form suitable for L'Hôpital's Rule. Another participant suggests taking the natural logarithm of the expression to facilitate the limit evaluation. There is also a discussion about the correct interpretation of the expression, with questions raised about whether it should be read as \(x^{x^2}\) or \((x^x)^2\).

Discussion Status

Participants are actively engaging with the problem, with some providing insights on how to approach the limit using logarithms and L'Hôpital's Rule. There is a recognition of the need for clarity in interpreting the expression, and some participants indicate a better understanding as the discussion progresses.

Contextual Notes

There is a focus on ensuring the correct interpretation of the limit expression, which is crucial for applying the appropriate mathematical techniques. The original poster's uncertainty highlights the importance of clear notation in mathematical discussions.

laker88116
Messages
57
Reaction score
0
[tex]\lim x \rightarrow 0+[/tex] [tex]x^x^2[/tex]. This should read x^x^2 if that isn't clear. I am not sure where to start. I know that I need to use l'hospital's Rule and I know that I need to get this in a form of [itex]\frac{f(x)}{g(x)}[/itex]. Any suggestions on what else to do?
 
Last edited:
Physics news on Phys.org
Hmm, if you take logs of both sides, what could you get?
Let [tex]y = \lim_{x \rightarrow 0 ^ +} x^{(x ^ 2)}[/tex]
So:
[tex]\ln y = \lim_{x \rightarrow 0 ^ +} x ^ 2 \ln x = z[/tex]
Now, to find:[tex]\lim_{x \rightarrow 0 ^ +} x ^ 2 \ln x[/tex], you can use L'Hopital rule. Rewrite it as:
[tex]z = \lim_{x \rightarrow 0 ^ +} x ^ 2 \ln x = \lim_{x \rightarrow 0 ^ +} \frac{\ln x}{\frac{1}{x ^ 2}}[/tex]
So [tex]y = e ^ z[/tex].
Viet Dao,
 
But how do you know that its isn't [tex](x^x)^2[/tex]? I understand what you are saying though and it makes perfect sense, I just wasn't sure where to break it up.
 
Because [tex](x^x)^2 = x^{2x}[/tex] and it would seem silly to write it that way.
 
Ok, well I understand now I was able to interpret it after VietDao29 did most of the work, but anyways, thanks for the help I see it now.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
6
Views
3K
Replies
12
Views
2K
Replies
17
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K