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I would really appreciate if someone could help me figure out where to start on this problem.

The question is to determine if $f$ is continuous at the origin.

$$\begin{equation} f(x,y)=\begin{cases} \dfrac{\sin(x-y)}{|x| + |y|}, & \text{if $(x,y) \neq (0,0)$}.\\ 0, & (x,y) = (0,0). \end{cases} \end{equation} $$

I'm really struggling with these concepts and would appreciate if someone could help me out

Ivo Terek
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manny
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3 Answers3

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A necessary condition for that function to be continuous at the origin is that the limit of the function along any path to the origin must equal $f(0,0) = 0$.

See what happens if you approach along an axis.

Kaj Hansen
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  • I don't have enough reputation to vote this up but thank you – manny Jul 15 '14 at 03:36
  • @manny, I'm glad I could help! For now, I believe you are allowed to "accept" the most helpful answer (by clicking the green arrow). This is good practice on Math:SE, and it lets future users know your question has been answered. – Kaj Hansen Jul 15 '14 at 03:39
  • Okay! Actually @Kaj Hansen, can you please help me understand Surb's solution if you have the time? I don't understand why Surb chose to drop the absolute value bars and why the one-sided limits are different. – manny Jul 15 '14 at 03:45
  • @manny, sure thing. In his original response, he dropped the absolute values because he is approaching $0$ from the positive side. So when $x>0$, $|x| = x$. From there, the limit can be evaluated using L'Hopital's rule. Next, if we consider $\lim_{x \rightarrow 0^-} \frac{\sin(x)}{|x|}$, then note it is the same as $\lim_{x \rightarrow 0^+}\frac{\sin(-x)}{x}$. If you again apply L'Hopital's rule to this, you'll find that it comes out to $-1$. – Kaj Hansen Jul 15 '14 at 03:51
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You can use L'hopital's rule to show that $\lim_{x-y \to 0} \sin(x-y)/(x-y) = 1$. Thus your limit is equivalent to $\lim_{x,y \to 0} \frac{x-y}{|x| + |y|}$. If you consider any line $y = ax$ and take the limit as $x \to 0$, you will get that the limit is not $0$. So the function is not continuous at $0$.

user2566092
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  • Thank you! Your approach is very interesting. In the future, how would I know to try something like that? @user2566092 – manny Jul 15 '14 at 03:39
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$$\lim_{x \to 0^+} f(x,0) = \lim_{x \to 0^+} \frac{\sin(x)}{x} = 1 \neq 0=f(0,0)$$ It follows that $f$ is not continuous at $(0,0)$.

Surb
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    Actually $\lim_{x\to 0-}\frac{\sin x}{|x|}=-1.$ – mfl Jul 13 '14 at 23:07
  • @mfl Right... I edited. – Surb Jul 14 '14 at 08:32
  • When you did the two sided limit, why did you drop the absolute value bars? I just don't understand why the one-sided limits are different. Thank you so much for your response though @Surb – manny Jul 15 '14 at 03:35
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    @manny It is not a two sided limit (but it is enough to show that one side already don't works). Here $x$ is approached $0$ by the right, i.e. $x > 0$ and $x \to 0$. Maybe you are more used to this notation: $\lim_{x \underset{>}{\to} 0} f(x,0)$. – Surb Jul 15 '14 at 07:50