Limit of two-variable function

  • Thread starter Thread starter twoflower
  • Start date Start date
  • Tags Tags
    Function Limit
Click For Summary
SUMMARY

The discussion focuses on evaluating the limit of the two-variable function \(\lim_{[x,y] \rightarrow [0,0]} (x^2+y^2)^{xy}\). The approach involves rewriting the limit using the exponential function: \(\lim_{[x,y] \rightarrow [0,0]} \exp^{xy \log (x^2+y^2)}\). The key conclusion is that \(\lim_{t \rightarrow 0+} t \log t = 0\) can be proven using L'Hôpital's rule by transforming it into the form \(\frac{\log{t}}{\frac{1}{t}}\), which is an indeterminate form of \(\frac{\infty}{\infty}\).

PREREQUISITES
  • Understanding of limits in multivariable calculus
  • Familiarity with exponential functions and logarithms
  • Knowledge of L'Hôpital's rule for evaluating indeterminate forms
  • Basic algebraic manipulation skills for rearranging equations
NEXT STEPS
  • Study the application of L'Hôpital's rule in detail
  • Learn about the properties of logarithmic functions in calculus
  • Explore multivariable limit evaluation techniques
  • Investigate the behavior of functions approaching zero in calculus
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus and analysis, as well as educators looking for effective methods to teach limit evaluation techniques.

twoflower
Messages
363
Reaction score
0
Hi all,

suppose I want to get this:

[tex] \lim_{[x,y] \rightarrow [0,0]} (x^2+y^2)^{xy}[/tex]

Here's how I approached:

[tex] \lim_{[x,y] \rightarrow [0,0]} (x^2+y^2)^{xy} = \lim_{[x,y] \rightarrow [0,0]} \exp^{xy \log (x^2+y^2)}<br /> <br /> \lim_{[x,y] \rightarrow [0,0]} xy \log (x^2 + y^2) = (x^2 + y^2) \log (x^2 + y^2) \frac{xy}{x^2 + y^2} \rightarrow 0[/tex]

Because the last fraction is bounded and the part before it goes to 0 (I hope).

But that's the problem, I don't know how to prove

[tex] \lim_{t \rightarrow 0+} t\ \log t = 0[/tex]

Thank you for help.
 
Physics news on Phys.org
You can use L'Hopital rule to prove that. Try rearrange the equation to:
[tex]\lim_{t \rightarrow 0 ^ +} t \log{t} = \lim_{t \rightarrow 0 ^ +} \frac{\log{t}}{\frac{1}{t}}[/tex]. Now it's in form [tex]\frac{\infty}{\infty}[/tex]. Can you go from here?
Viet Dao,
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
3K
  • · Replies 8 ·
Replies
8
Views
1K
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 23 ·
Replies
23
Views
3K