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$$f(x,y) = \begin{cases} \dfrac{\sin(xy)}{xy} & \text{if $x y \ne 0$} \\ 1 & \text{if $xy=0$} \end{cases}$$ all ideas are appreciated

i think this is non-continuous, i did by converting to polar coordinates

Looking for more ideas and interesting observations

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Please do not use displaystyle math in titles. –  AlexR Aug 14 at 8:30
    
@triple_sec do not alter the mathematical meaning of expressions without asking the OP. –  AlexR Aug 14 at 8:32
    
@AlexR It looks like an obvious omission, though... –  triple_sec Aug 14 at 8:33
    
@triple_sec Still the OP should at least show that much effort into a question to confirm this. Also, $y$ may also be an arbitrary constant. –  AlexR Aug 14 at 8:37
    
the function is $f(x, y)$, not $f(x)$ –  loading.... Aug 14 at 8:48

2 Answers 2

I assume that $f(x)$ should read as $f(x,y)$. Define the function $g:\mathbb R^2\to\mathbb R$ as $g(x,y)=xy$ for all $(x,y)\in\mathbb R^2$. This function is clearly continuous. Moreover, define another function $h:\mathbb R\to\mathbb R$ as \begin{align*}h(z)=\begin{cases}\dfrac{\sin z}{z}&\text{if $z\neq0$,}\\1&\text{if $z=0$.}\end{cases}\end{align*} This function is continuous, as it is well-known that $\lim_{z\to0}(\sin z)/z=1$. Now, observe that $f=h\circ g$ and recall that the composition of continuous functions is continuous.

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What does $R^2 -> R$ mean? The pairs of real numbers being mapped to real numbers? That doesn't make sense. –  Celeritas Aug 14 at 10:14
    
@Cthulhu I think we have a misunderstanding, when I ask what it means, it doesn't help when reply "what do you think it should be"? –  Celeritas Aug 14 at 17:02
    
@Celeritas $g:\mathbb R^2\to\mathbb R$ means that to each pair of real numbers $(x,y)\in\mathbb R^2$, the function $g$ assigns a unique real number, denoted as $g(x,y)$. In this special case, this real number which is assigned to $(x,y)$ is the product $xy$. –  triple_sec Aug 14 at 19:16

If $y$ is presumed constant (as you write $f(x)$), the function is indeed continuous wich can be proven by showing that $$\lim_{x\to0} \frac{\sin x}x = 1$$ (or trivially if $y=0$) And noting that $f = x\mapsto \frac{\sin x}x \circ x\mapsto xy$.
If $y$ is also a variable (i.e. you meant $f(x,y)$) take a look if $(x,y) \mapsto xy$ is continuous (yes, it is) and note again that $$f = x\mapsto \frac{\sin x}x \circ (x,y) \mapsto xy$$ is continuous as the composition of two continuous functions.

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