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Let $x=r\cos(\theta)$ et $y=r\sin(\theta)$ and $f(0,0)=0$. Suppose I've shown that $$ \left|f\left(x,y\right)-f\left(0,0\right) \right| \leq \left|\sin^3\left(\theta\right)\right| $$ Can I conclude that $f$ is continuous ? Meaning, does I have $$\sin^3\left(\theta\right)\underset{(x,y) \rightarrow 0}{\rightarrow}0 \ ?$$

Atmos
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  • No. ${}{}{}{}{}$ –  Apr 28 '18 at 15:59
  • @T.Bongers how do you do that? Whenever "No." is enough I struggle to find 12 additional stupid characters... – Arnaud Mortier Apr 28 '18 at 16:03
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    @ArnaudMortier Put a bunch of braces {}{}{} inside dollar signs. MathJax will interpret it as empty space but it meets the comment requirement. –  Apr 28 '18 at 16:05

2 Answers2

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You can have $(x,y)$ tend to $(0,0)$ just by making $r$ smaller and smaller without acting on $\theta$. So the answer is no - more precisely, $\sin^3 \theta$ does not have a limit at all as $(x,y)\to (0,0)$.

Arnaud Mortier
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  • So I can conclude that it is not continuous on $(0,0)$ ? – Atmos Apr 28 '18 at 16:09
  • @Atmos no, with just the data given you cannot conclude anything. $f$ could simply be identically $0$ and it would then be continuous. Or it could be $\sin ^3 \theta$ itself and then not be continuous. – Arnaud Mortier Apr 28 '18 at 16:10
  • In fact, i've $f\left(x,y\right)=\frac{y\left(x^2+y^2\right)\cos\left(xy\right)-x\sin\left(x,y\right)}{\left(x^2+y^2\right)^{3/2}}$. So $f \ne 0$. – Atmos Apr 28 '18 at 16:12
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    I didn't think that your particular $f$ is zero. That was just to illustrate that anything can happen with the given data. – Arnaud Mortier Apr 28 '18 at 16:13
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Actually, you cannot conclude that $f(x,y) \to 0$, Since the limit of $f(x,y)$ relied on the $\theta$. Here is a counterexample:

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

$\lim_{(x,y)\to 0} f(x,y) = \cos \theta \sin \theta $, the limit is rely on the $\theta$ even though $r$ is vanish.