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I have some trouble with this limit:

$$\lim_{(x,y)\to (0,0)} \frac{x^3+y^2}{x^2+|y|}.$$

My first attempt to solve it was with polar coordinates but I couldn't find an expression which was independent of $\varphi$. Now I'm trying to solve it with the triangle inequality so I can find an upper limit that squeezes my expression into $0$. My work so far:

$$\left| \frac{x^3+y^2}{x^2+|y|} \right| \leq \frac{|x^3+y^2|}{|x^2+|y||} \leq \frac{|x^3|+|y^2|}{|x^2|+||y||} = \frac{|x^3|+y^2}{x^2+|y|}.$$

But from here I don't know how to continue.

Thanks in advance!

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  • $\begingroup$ Just a comment: the step $\dfrac{|x^3+y^2|}{|x^2+|y||}\le\dfrac{|x^3|+|y^2|}{|x^2|+|y|}$ is invalid, because you're making BOTH the numerator and the denominator bigger - the result is ambiguous. If you make the numerator bigger and the denominator smaller (to a point), then the fraction as a whole gets bigger. $\endgroup$ – Adrian Keister May 9 '18 at 13:27
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By Cauchy–Schwarz

$$|x^3+y^2|=|xx^2+|y||y||\le\sqrt{x^2+y^2}\sqrt{x^4+y^2}\le \sqrt{x^2+y^2}\sqrt{(x^2+|y|)^2}=\sqrt{x^2+y^2}(x^2+|y|)$$

then

$$\left|\frac{x^3+y^2}{x^2+|y|}\right|\le\sqrt{x^2+y^2}\to 0 \implies \frac{x^3+y^2}{x^2+|y|}\to 0$$

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$\left|\frac{x^3+y^2}{x^2+|y|}\right|\leq \frac{|x^3|+|y^2|}{x^2+|y|}=\frac{|x^3|}{x^2+|y|}+\frac{|y^2|}{x^2+|y|}\leq \frac{|x^3|}{x^2}+\frac{|y^2|}{|y|}=|x|+|y|$

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