# Total differential of a function

Let's consider a function $$f(x,y)=\begin{cases}\dfrac{xy^3}{\sqrt{x^2+y^2}},& (x,y)\neq(0,0)\\ 0,& (x,y)=(0,0)\end{cases}$$ does it have a total differential in point $(0,0)$? I say that it does not, since I believe, that the limit does not exist. However, I was proven wrong by wolfram alpha, who says that it does exist and is $0$. I'd like to ask how to prove the limit exists.

Another method: Try $\frac{xy^3}{x^2+y^2} \leq \frac{x^4+y^4}{x^2+y^2} \leq \frac{(x^2+y^2)^2}{x^2+y^2} \leq x^2+y^2 \rightarrow 0$. Is this sufficient? If so, why?

• Yes, because $0\leq xy\leq\frac{xy^3}{x^+y^}\leq x^2+y^2$ because of two policemen theorem and we know that $\lim_{x,y\to0,0 }x^2+y^2=\lim_{xy}xy=0$ – user74200 Jun 29 '14 at 15:50

Use polar coordinates: $x = r\cos(\theta)$, $y = r\sin(\theta)$. Then \begin{align} f(r, \theta) = r^{3}\cos(\theta)\sin(\theta)^{3}, \end{align} which show that the singularity at $(x,y) = (0,0)$ is removable. Differentiating with respect to $r$ or $\theta$ gives a function continuous at $(0,0)$.

• You have a typo in the exponent of $\sin \theta.$ – mfl Jun 23 '14 at 0:12
• However, a function $f(x,y)=xy/(x^2+y^2)^{1/2}$ also grants us, that the singularity disappears, but it does not have limit in (0,0). – user74200 Jun 23 '14 at 0:29
• Good catch. The coordinate transformation on @user74200's function would give $r\sin(2\theta)/2$, which when differentiated wrt $r$ would clearly show that the limit does not exist. – user14717 Jun 23 '14 at 0:40

Using $$\frac{xy}{x^2+y^2}\leq \frac{1}{2}$$ we see that $$\left|\frac{xy^3}{\sqrt{x^2+y^2}}\right|\leq \frac{1}{\sqrt{2}} |x|^{\frac{1}{2}}|y|^{\frac{5}{2}}$$ and so we see the limit at the origin is zero.

Only @ChrisK is addressing your original question. They have argued that $f$ is continuous at $0$. Now you compute that both partial derivatives of $f$ at $0$ are $0$ and so the best linear approximation — if it exists — is the $0$ map. Now you must check that $$\lim_{(x,y)\to (0,0}\frac{f(x,y)-f(0,0)-(0x+0y)}{\sqrt{x^2+y^2}}=0.$$ But $$\left|\frac{xy^3}{x^2+y^2}\right|\le \frac{(\sqrt{x^2+y^2})^4}{x^2+y^2}=x^2+y^2\to 0,$$ as required.

• Can you please explain how you reach the desired upperbound $\left|\frac{xy^3}{x^2+y^2}\right|\le \frac{(\sqrt{x^2+y^2})^4}{x^2+y^2}$ ? You referenced Chris' (inactive user) answer for this but I don't understand how we get that $xy^{3}\leq x^{4}+y^{4}$. Thank you so much in advance. – John11 Jun 6 '17 at 10:27
• @John11: We have $|x|,|y|\le \sqrt{x^2+y^2}$. – Ted Shifrin Jun 6 '17 at 15:30