Finding the limit of a multivariable function

  • #1
schniefen
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Homework Statement:
Find the limit of the function below as ##\textbf{x}\to\textbf{0}##.
Relevant Equations:
##\frac{e^{|\textbf{x}|^2}-1}{|\textbf{x}|^2+x^2_1x_2+x^2_2x_3}## where ##\textbf{x}=(x_1,x_2,x_3)## and ##|\textbf{x}|=\sqrt{x^2_1+x^2_2+x^2_3}##.
If one approaches the origin from where ##x_2=0##, the terms ##x^2_1x_2+x^2_2x_3## in the denominator equal ##0##. Substituting ##|\textbf{x}|^2## for ##t## yields the expression ##\frac{e^t-1}{t}##, which has limit 1 as ##\textbf{x}\to\textbf{0}## and thus ##t\to0##. So the limit should be 1 if it exists. How could one show that it does exist?
 

Answers and Replies

  • #2
BvU
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How could one show that it does exist?
You need to prove that, no matter how you approach the origin, the limit will always be the same...
Try a few candidates that might yield something different, to get a feeling.
 
  • #3
schniefen
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You need to prove that, no matter how you approach the origin, the limit will always be the same...
Try a few candidates that might yield something different, to get a feeling.

Yes, but approaching along ##x_1=0## or ##x_3=0## doesn't simplify the expression, does it?
 
  • #4
BvU
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Oh, doesn't it :rolleyes: ?
What comes out ?
 
  • #5
schniefen
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It only cancels one of the terms in the sum ##x^2_1x_2+x^2_2x_3## for ##x_1=0## and ##x_3=0## respectively.
 
  • #6
BvU
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And what is left over ?

Tip: substitute ## | x| = \sqrt {...} ## to get a clearer picture, especially in the denominator 😉 .
 
  • #7
BvU
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Bedtime here (UTC+2) :sleep:
 
  • #8
HallsofIvy
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With a multi-variable limit, [itex]\left(x_1, x_2, x_3\right)[/itex] going to (0, 0, 0), it might be best to convert to spherical coordinates. That way the limit is just as [itex]\rho[/itex] goes to 0. If that limit depends on [itex]\theta[/itex] or [itex]\phi[/itex] there is no limit. If it is a single number, then that is the limit.
 
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