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Use the symbols [tex]\beta[/tex] and [tex]\sigma[/tex] definition of limit to prove that limit (x,y)[tex]\Longrightarrow[/tex](0,0)x+y/x[tex]\x^2[/tex]+y[tex]\y^2[/tex]=0

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- Thread starter kidia
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- #1

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Use the symbols [tex]\beta[/tex] and [tex]\sigma[/tex] definition of limit to prove that limit (x,y)[tex]\Longrightarrow[/tex](0,0)x+y/x[tex]\x^2[/tex]+y[tex]\y^2[/tex]=0

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EnumaElish

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What is a beta and sigma limit definition? Same as the epsilon and delta definition? Here are some examples:

http://archives.math.utk.edu/visual.calculus/1/definition.6/

http://omega.albany.edu:8008/calc3/several-vars-dir/lim-epsilon-delta-m2h.html

http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preciselimdirectory/PreciseLimit.html

http://archives.math.utk.edu/visual.calculus/1/definition.6/

http://omega.albany.edu:8008/calc3/several-vars-dir/lim-epsilon-delta-m2h.html

http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preciselimdirectory/PreciseLimit.html

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rachmaninoff

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quasar987

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[tex]\frac{x+y}{x^2+y^2}<\beta[/itex]

So we kinda want to find a function [itex]\sigma(\beta)[/itex].

The statement "the distance from the origin of the point (x,y) is smaller than [itex]\sigma[/itex]" is written mathematically as [itex]\sqrt{x^2+y^2}<\sigma[/itex]

There are many solutions but here's a hint based on one:

Use the fact that [itex]x+y \leq (x^2+y^2)^2[/itex] coupled with the hypothesis [itex]\sqrt{x^2+y^2}<\sigma[/itex] to define a function [itex]\sigma(\beta)[/itex].

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