Prove the function is continuous

User16

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Need to prove this function is continuous f(x,y)=sin(xy)/sqrt(x^2 + y^2) [ analytically ]
Now |f(x,y)-f(a,b)|
<=|f(x,y)|+|f(a,b)|
<=|1/sqrt(x^2 + y^2)| + |1/sqrt(a^2 + b^2)|
<= 1/|x| + 1/|a|
= (|x|+|a|)/|a||x|
<=( x^2 + a^2)/|a||x| .
I am getting stuck at this position . Any ideas ? . Consider any neighborhood circular or square.
 
I wouldn't approach it that way. Since the function is the quotient of continuous functions it is continuous if the denominator is not zero so all we need to worry about is (0,0).
Now
x2+y2x2=x\displaystyle \sqrt{x^2 + y^2} \ge \sqrt{x^2} = |x|.
and, if x is not zero
1x2+y21x\displaystyle \frac{1}{\sqrt{x^2 + y^2}} \le \frac{1}{|x|}.
If x is zero, consider the same type thing with y. Thus
sin(xy)x2+y2sin(xy)x= ysin(xy)xy \displaystyle |\frac{sin(xy)}{\sqrt{x^2 + y^2}}| \le |\frac{sin(xy)}{x}| = |\space \frac {y sin(xy)}{xy}\space |.

Now let x and y go to zero. Depending on just what is wanted, you would have to play around a bit more for an epsilon-delta proof.
 
That function quite clearly is NOT continuous at (0, 0) because it has no value there. Did you mean that
f(x)=sin(xy)x2+y2\displaystyle f(x)= \frac{sin(xy)}{\sqrt{x^2+ y^2}} if (x, y) is not (0, 0), while f(0, 0)= 0?
 
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