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The limit is $$\lim_{(x,y)\to(0,0)}\frac{x^3y}{x^4+y^2}$$ As usual, I tried checking along particular paths, namely the axes and the curves $y=mx^n$ for various values of $n$, but to no avail; all the limits evaluate to $0$. I resorted to converting to polar coordinates, giving $$\lim_{r\to0^+}\frac{r^4\cos^3t\sin t}{r^2\left(r^2\cos^4t+\sin^2t\right)}$$ In an attempt to apply the squeeze theorem, I found that trig expression in the numerator is bounded, with $|\cos^3t\sin t|\le\dfrac{3\sqrt3}{16}$, and that in the denominator I have $$\begin{align*} |r^2\cos^4t+\sin^2t|&\le r^2|\cos^4t|+|\sin^2t|\\ &\le r^2+1 \end{align*}$$ and so $$\lim_{r\to0^+}\frac{r^4\cos^3t\sin t}{r^2\left(r^2\cos^4t+\sin^2t\right)}=\frac{3\sqrt3}{16}\lim_{r\to0^+}\frac{r^4}{r^4+r^2}=0$$ Is my reasoning valid? I am unsure about whether or not the bound for the denominator is correct.

user170231
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  • No: you must find a lower bound for the denominator in order to have an upper bound for the fraction. – Bernard Jan 06 '15 at 17:22
  • How do you get your last equality from your trigonometric inequalities? – Martigan Jan 06 '15 at 17:23
  • @Bernard if I were to find that the lower bound is $1$, would I then be able to say $|r^2\cos^4t+\sin^2t|\ge1$ (which I found with Mathematica) means the limit is equivalent to $\dfrac{3\sqrt3}{16}\lim\limits_{r\to0^+}\dfrac{r^4}{r^2}$? – user170231 Jan 06 '15 at 18:01

1 Answers1

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$$ (x^2 - y)^2 \geq 0, $$ $$ x^4 - 2 x^2 y + y^2 \geq 0, $$ $$ x^4 + y^2 \geq 2 x^2 y. $$

$$ (x^2 + y)^2 \geq 0, $$ $$ x^4 + 2 x^2 y + y^2 \geq 0, $$ $$ x^4 + y^2 \geq -2 x^2 y. $$

$$ x^4 + y^2 \geq 2 x^2 |y|. $$

$$ \color{magenta}{ \left| \frac{x^2 y}{x^4 + y^2} \right| \leq \frac{1}{2} } $$

Your numerator is $x^3 y$ rather than $x^2 y;$ what happens now?

Some detail, not directly needed for your question but probably helpful anyway, at recent Continuity in $\mathbb R^n$

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Will Jagy
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