Given a function defined by

$$ f(x,y):=\begin{cases} \large\frac{2x^{5}-3y^{5}}{\left(x^{2}+y^{2}\right)^{2}} & (x,y) \neq(0,0) \\ 0 & (x,y)=(0,0) \end{cases}$$ Show that the function is continuous at the origin and its partial derivatives of the first order with respect to each one of the independent variables exist,but it's not differentiable at the origin.

The function is continuous at the origin iff $\lim_{\left(x,y\right) \to \left(0,0\right)}\frac{2x^{5}-3y^{5}}{\left(x^{2}+y^{2}\right)^{2}}$ has a finite value and is equal to $0$.

So we have:

$$\lim_{\left(x,y\right) \to \left(0,0\right)}\frac{2x^{5}-3y^{5}}{\left(x^{2}+y^{2}\right)^{2}}=\lim_{r \to 0}\frac{r^{5}\left(2\cos^{5}\left(\theta\right)-3\sin^{5}\left(\theta\right)\right)}{r^{4}}=0$$

Which implies the function is continuous at $(0,0)$.

On the other hand: $$\frac{\partial f(x,y)}{\partial x}\Big|_{(0,0)} \;\;\text{does exist}\;\; \iff \lim_{h \to 0}\frac{f\left(0+h,0\right)-f\left(0,0\right)}{h} \;\;\text{does exist}\;\;$$

Which does,and is equal to $2$.

Also $$\frac{\partial f(x,y)}{\partial y}\Big|_{(0,0)} \;\;\text{does exist}\;\; \iff \lim_{h \to 0}\frac{f\left(0,0+h\right)-f\left(0,0\right)}{h} \;\;\text{does exist}\;\;$$Which does,and is equal to $-3$.

Gathering these all implies that the partial derivatives of the function of the first order with respect to each one of the variable exist.

The derivative of the function at the origin exist iff the following limit exists:

$$\lim_{\left(x,y\right) \to \left(0,0\right)}\frac{f\left(x,y\right)-f\left(0,0\right)}{\sqrt{x^{2}+y^{2}}}$$

The limit can be written as: $$\lim_{\left(x,y\right) \to \left(0,0\right)}\frac{r^{5}\left(2\cos^{5}\left(\theta\right)-3\sin^{5}\left(\theta\right)\right)}{r^{5}}=2\cos^{5}\left(\theta\right)-3\sin^{5}\left(\theta\right)$$

Which is not unique for different $\theta$s,implies the function is not differentiable at the origin.

Another approach would be setting $y=0$ and computing the limit:

$$\lim_{x \to 0}\frac{2x^{5}}{x^{4}\left|x\right|}$$

Depending on $\lim_{x \to 0^+}$ and $\lim_{x \to 0^-}$ the limit alternates between $2$ and $-2$,which is not unique,implies the limit does not exist at $(0,0)$.

The problem is that I used the polar coordinates,but I'm not sure if I'm allowed to do that,since I don't know when we are able to use this trick and be sure that the answer is the real answer.

Besides can someone please check the solution and give me another version without using polar coordinates (and explain when we are allowed to use polar coordinates).

  • $\begingroup$ Derivative exists if the limit $\lim_{\left(x,y\right) \to \left(0,0\right)}\frac{f\left(x,y\right)-f\left(0,0\right)-x\frac{\partial f(x, y)} {\partial x}\Big |_{(0,0)}-y\frac{\partial f(x, y) }{\partial y}\Big |_{(0,0)}} {\sqrt{x^{2}+y^{2}}} $ exists. $\endgroup$ – Koro Jun 20 '20 at 11:02

Polar coordinates is just change of coordinates so if limit exists in one system, it does in the other also.
Now to prove the existence of limit without using polar coordinates:
First of all, I claim that limit is 0. By triangular inequality we have, $|2x^5-3y^5|\le 2|x^5|+3|y^5|$. Now noting that, $(x^2+y^2)^2\ge (x^2)^2$ and also, $\ge (y^2)^2$
Therefore for $(x, y) \ne (0,0)$, we have, $\frac{|2x^5-3y^5|}{(x^2+y^2)^2}\le\frac{2|x^5|}{(x^2+y^2)^2}+\frac{3|y^5|}{(x^2+y^2)^2}\le2|x|+3|y|\lt2\sqrt{x^2+y^2}+3\sqrt{x^2+y^2}=5\sqrt{x^2+y^2}\lt \epsilon $,where $\epsilon$ is any positive number. Now choose, $\delta =\epsilon /5$ and by $\epsilon - \delta$ definition of limit, it follows that $\lim_{\left(x,y\right) \to \left(0,0\right)}\frac{2x^{5}-3y^{5}}{\left(x^{2}+y^{2}\right)^{2}}=0$
Note: Please note that, as I mentioned in my comment, derivative exists if the following limit exists: $\lim_{\left(x,y\right) \to \left(0,0\right)}\frac{f\left(x,y\right)-f\left(0,0\right)-x\frac{\partial f(x, y)} {\partial x}\Big |_{(0,0)}-y\frac{\partial f(x, y) }{\partial y}\Big |_{(0,0)}} {\sqrt{x^{2}+y^{2}}}$


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