1
$\begingroup$

Let $f: \mathbb{R}^2 \to \mathbb{R}$ be given by

$$ f(x,y) = \begin{cases} \frac{\sin(xy)}{\sqrt{x^2+y^2}}, & \text{if }(x,y)\text{ $\neq (0,0)$} \\ 0, & \text{if }(x,y)\text{ $= (0,0)$} \end{cases} $$

My Attempt:

Notice clearly $f$ is continuous at any point in $\mathbb{R}^2 \setminus \{(0,0) \} $ since $\sin \alpha$, polynomials, and division of continuous functions are again continuous. To study the continuity at the origin, consider the sequence $\mathbf{x_n} = ( \frac{1}{n}, \frac{1}{n} ) $. Notice $\mathbf{x_n} \to (0,0) $. However,

$$ f( \mathbf{x_n}) = \frac{\sin( \frac{1}{n^2})}{\frac{ \sqrt{2}}{n}} $$

How can I show that this limit does not tend to $0$ ?

$\endgroup$
2
  • 1
    $\begingroup$ But it does tend to $0$. $\endgroup$ Feb 19 '14 at 10:04
  • $\begingroup$ Remember that $\displaystyle \frac{\sin(\frac{1}{n^2})}{\frac{1}{n^2}} \to 1$ as $n \to \infty$ $\endgroup$
    – dani_s
    Feb 19 '14 at 10:12
3
$\begingroup$

Use polar coordinates $x=r\cos(\varphi)$, $y=r\sin(\varphi)$. Now $(x,y)\rightarrow (0,0)$ if and only if $r\rightarrow 0$. Then

$$\frac{\sin(xy)}{\sqrt{x^2+y^2}}=\frac{1}{r}\sin(r^2\cos(\varphi)\sin(\varphi))=r\cos(\varphi)\sin(\varphi)+O(r^5)\longrightarrow 0$$

as $r\rightarrow 0$, because $|r\cos(\varphi)\sin(\varphi)|\le r$. So your function is continuous at $(0,0)$.

Edit:

Here as requested the version without using polar coordinates:

We have $\sqrt{x^2+y^2}\ge |x|+|y|$ and $\sin(xy)\le |xy|$ for $|xy|$ small, so

$$\left|\frac{\sin(xy)}{\sqrt{x^2+y^2}}\right|\le \frac{|xy|}{|x|+|y|}$$

If $x=0$ or $y=0$ this is $0$. For $x,y\not=0$ we have

$$\frac{|xy|}{|x|+|y|}=\frac{1}{\frac{1}{|x|}+\frac{1}{|y|}}$$

which converges to $0$ for $x,y\rightarrow 0$.

$\endgroup$
2
  • $\begingroup$ is there another way to show that the limit is $0$ ? without using polar coordinates ? $\endgroup$
    – user124140
    Feb 19 '14 at 11:51
  • $\begingroup$ @Learner: Yes, see above. $\endgroup$
    – J.R.
    Feb 19 '14 at 12:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy