Show continuity at $(0,0)$ of $f(x,y)=\frac{xy}{\sqrt{x^2+y^2}}$ for $(x,y)\neq (0,0)$ and $f(0,0)=0$

Question

Show continuity at $(0,0)$ of $f(x,y)=\frac{xy}{\sqrt{x^2+y^2}}$ for $(x,y)\neq (0,0)$ and $f(0,0)=0$.

A solution I saw was to write

$$\lim_{(x,y)\to 0}x\dfrac{y}{\sqrt{x^2+y^2}}$$

$$\lim_{(x,y)\to 0}x \cdot \lim_{(x,y)\to 0}\dfrac{y}{\sqrt{x^2+y^2}}$$

The term on the left tends to $0$ and the term on the right has absolute value less than $1$.

My Question

Originally I tried to use polar coordinates to solve.

$$f(x,y)=f(r\cos\theta,r\sin\theta)$$

$$f(x,y)=\dfrac{xy}{\sqrt{x^2+y^2}}= r\cos\theta \sin\theta$$

But I wasn't sure how to proceed. I'm not comfortable working in polar form but I'm thinking it's obvious that $r\cos\theta\sin\theta \to 0$ as $(r\cos\theta,r\sin\theta)$.

Is it the case that for $(r\cos\theta,r\sin\theta) \to (0,0)$ we must have that $r \to 0$ since when $\sin\theta=0 \Rightarrow \cos\theta \neq 0$.

Answer 1

If you are not comfortable with polar coordinates, your remark

"...and the term on the right has absolute value less than 1"

suffices to show that $\left|x \frac{y}{\sqrt{x^2+y^2}} \right|\leq \left| x \right| \leq \sqrt{x^2+y^2}$ and thus $f(x,y) \rightarrow0$ as $(x,y)\rightarrow 0$.

Answer 2

You can use $|\sin\theta|\leq1$ and $|\cos\theta|\leq1$ to make $$r\sin\theta\cos\theta\to0$$ as $r\to0$. Like Descartes coordinates here we have both $r\to0$ and $\theta\to0$ for evaluating the limit. Then $$\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{(r,\theta)\to(0,0)}r\cos\theta\sin\theta \to0$$ means the function is continuous in origin.

Answer 3

Your argument is correct: $(r \cos \theta, r \sin \theta) = (0,0)$ holds if and only if $r = 0$ , so approaching the origin in polar coordinates indeed corresponds to letting $r \to 0$ . You can, however, approach the origin from different directions, which is something you need to be careful about when working in polar coordinates! The limit $ \lim_{(x,y) \to (0,0)} f(x,y)$ exists if and only if the limits $$ \lim_{r \to 0} f(r \cos \vartheta(r), r \sin \vartheta(r))$$ exist and agree for every parametrisation $\vartheta: (0,1) \to [0,2\pi)$ of the polar angle.

In your problem this works out: $$ \lim_{r\to 0} f(r \cos \vartheta(r), r \sin \vartheta(r)) = \lim_{r\to 0} \, r \cos \vartheta(r) \sin \vartheta(r) = 0$$ holds for every $\vartheta$ since $\cos$ and $\sin$ are bounded by $1$ .

If we consider $g(x,y) = \frac{x}{\sqrt{x^2 + y^2}}$ for $(x,y) \neq (0,0)$ instead, things are quite different. We have $$ g(r \cos(\theta), r \sin(\theta)) = \cos \theta ,$$ so $$ \lim_{r \to 0} g(r \cos(0), r \sin(0)) = 1 \neq -1 = \lim_{r \to 0} g(r \cos(\pi), r \sin(\pi)) \, .$$ Therefore the limit $\lim_{(x,y) \to (0,0)} g(x,y)$ does not exist and we cannot choose a value for $g(0,0)$ in order to make $g$ continuous.

A better example of the issue is provided by this question. Consider $$ h(x,y) = \frac{x^2 y}{x^4 + y^2} = \frac{r \cos^2 \theta \sin \theta}{r^2 \cos^4 \theta + \sin^2 \theta} \, . $$ At first glance one would probably expect the limit $0$ at the origin, given the above representation in polar coordinates. However, along the parabolic path $y = x^2$ the limit is equal to $\frac{1}{2}$, so the limit $\lim_{(x,y) \to (0,0)}h(x,y)$ does not exist.

In polar coordinates this path is parametrised implicitly by $$\frac{\sin \vartheta(r)}{\cos^2 \vartheta(r)} = r \, $$ or explicitly by $$ \sin \vartheta(r) = \frac{\sqrt{1+4 r^2}-1}{2r} \, , $$ which is not easy to spot. This illustrates the fact that using polar coordinates is possible, but not always advisable when working on these problems.

Answer 4

We can also use that

$$\frac{|xy|}{\sqrt{x^2+y^2}}\le \frac{|xy|}{\sqrt{2}\sqrt{|xy|}}=\frac{1}{\sqrt{2}}|xy|^{1/2}$$