0

I am unable to find by definition the continuity of the following function $$ f(x,y)=\sin\left(\frac xy\right)$$ at $(0,0)$ I have substituted $$x=my$$ and found the limit to be non-existent and hence the function as discontinuous at $$(0,0)$$. Am I correct?

rah22ul
  • 21
  • What do you know about the continuity of $x/y$ ? –  Mar 20 '17 at 11:40
  • 1
    I substituted $$x=my$$ and found the limit to be non-existent. So I concluded that it is not continuous. However, I wanted to be sure. – rah22ul Mar 20 '17 at 11:41
  • 1
    @rah22ul That last commentary you should put in the body of your question, so that everybody knows what you have done and where you're having doubts. By the way, what you did is correct: that proves the limit doesn't exist as you get the constant function $;\sin m;$ . – DonAntonio Mar 20 '17 at 11:44

3 Answers3

3

I think you can look at those two parametric curves

  1. $(t, t)$ where $t\to 0$
  2. $(t, t^2)$ where $t\to 0$

to figure out the limit doesn't exist (could have commented but don't have enough reputation to do so)

OliOliver
  • 175
2

The function is not defined at $\langle0,0\rangle$ hence is not continuous at $\langle0,0\rangle$.

Another question is: can we expand the function by giving it a value at $\langle0,0\rangle$ in such a way that this expanded function is continuous at $\langle0,0\rangle$?

Again the answer is: "no".

The sequences $(\langle\frac1{n},\frac1{n}\rangle)_n$ and $(\langle\frac2{n},\frac1{n}\rangle)_n$ both converge to $\langle0,0\rangle$ but substituting them in $f(x,y)$ gives us two convergent sequences having distinct limits ($\sin1\neq\sin2$).

That contradicts continuity at $\langle0,0\rangle$ of any expansion of the function at $\langle0,0\rangle$.

drhab
  • 151,093
0

put $y=\frac{1}{n\pi}$ and again put $y=\frac{1}{n\pi+\frac{\pi}{2}}$

Clearly $y\longrightarrow‎ 0$ when $n\longrightarrow‎ \infty$ but $sin(\frac{x}{y})\nrightarrow‎ 0$.

d.y
  • 649