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I am getting confused with this epsilon delta proof of the limit for this particular case. Prove that $\lim_{(x,y)\to(0,0)} (2x^2+3y^2)=0.$

if the limit is equal to zero, then for any given positive epsilon there exists some positive delta, such that $0$

so in this case i have $|2x^2+3y^2|$

I cannot seem to find the way to express delta in epsilon terms I feel like i need to find the relationship between the given function and the $\sqrt{x^2+y^2}$

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Hint:

For all $x$, $x^2 \geq 0$, so

$$2x^2 + 3y^2 \leq 3x^2 + 3y^2 = 3(x^2 + y^2) = 3\left(\sqrt{x^2 + y^2}\right)^2 \left(\simeq 3 \delta^2 \right)$$

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