- #1
HF08
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[SOLVED] Help me evaluate this limit
Part a:
Show that f(x,y) = [tex]\frac{x^{4}+y^{4}}{x^{2}+y^{2}}[/tex]
as (x,y) -> (0,0)
Part b:
Similarily, show that f(x,y) = [tex]\frac{\sqrt{\left|xy\right|}}{\sqrt[3]{x^{2}+y^{2}}}[/tex] as (x,y)->(0,0)
Lets work with Part a, shall we?
I seemed to be getting a [tex]\frac{0}{0}[/tex] type limit. In prior situations and examples, I have seen clever substitutions or inequality relationships to aid us. In this spirit of such an approach I have thought it obvious to write it as:
f(x,y) = [tex]\frac{(x^{2})^{2}+(y^{2})^{2}}{x^{2}+y^{2}}[/tex]
Unfortunate for me is that this rather unoriginal form lends me nothing (at least to me). I have the sum of squares, not the difference of squares. I have thought of multiplying both the numerator and denominator by an indentity of some kind, but this might also be provide poor results? Can you please help me here?
As for part b, I am sure similar strategies are at work. I am also aware that part b looks acutely more difficult. Hence, please help me work on part a first. It isn't lack of effort, I assure you as my post here implies.
Thank you,
HF08
Part a:
Show that f(x,y) = [tex]\frac{x^{4}+y^{4}}{x^{2}+y^{2}}[/tex]
as (x,y) -> (0,0)
Part b:
Similarily, show that f(x,y) = [tex]\frac{\sqrt{\left|xy\right|}}{\sqrt[3]{x^{2}+y^{2}}}[/tex] as (x,y)->(0,0)
Lets work with Part a, shall we?
I seemed to be getting a [tex]\frac{0}{0}[/tex] type limit. In prior situations and examples, I have seen clever substitutions or inequality relationships to aid us. In this spirit of such an approach I have thought it obvious to write it as:
f(x,y) = [tex]\frac{(x^{2})^{2}+(y^{2})^{2}}{x^{2}+y^{2}}[/tex]
Unfortunate for me is that this rather unoriginal form lends me nothing (at least to me). I have the sum of squares, not the difference of squares. I have thought of multiplying both the numerator and denominator by an indentity of some kind, but this might also be provide poor results? Can you please help me here?
As for part b, I am sure similar strategies are at work. I am also aware that part b looks acutely more difficult. Hence, please help me work on part a first. It isn't lack of effort, I assure you as my post here implies.
Thank you,
HF08