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$f(x,y)=\frac{xy^3}{x^2 + 4y^2}$,

$(x,y)$ not eaqual to $(0,0)$;

use $\epsilon-\delta$ definition to show that $f(x,y)$ tends to $(0,0)$.

I'm unsure how to deal with fractions for, $\epsilon-\delta$ proof... could applying the proof separately for numerator and denominator; then later combining them help?

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1 Answer 1

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$\mathbb x^2 + 4y^2\geq 4xy$ applying AM $\geq$ GM then $ f(x.y) \leq \frac{y^2}{4} $ from there you can get the $\epsilon-\delta $ proof

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so i essentially apply epsilon-delta to y^2/4. meaning the answer is delta< (4*epsilon)^0.5 –  redrum Oct 23 '12 at 13:23
    
oh and...is AM = arithmetic mean, and GM = geometric mean. Also your initial equation doesn't really remind me of how it links to AM or GM. –  redrum Oct 23 '12 at 13:40
    
yes AM = arithmetic mean, and GM = geometric mean from this (a+b)/2 >= sqrt(ab) take a=x^2 (>=0) and b= 4y^2(>=0) –  jim Oct 23 '12 at 14:10

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