What is the lime as (x,y) approaches (0,0) of (x^2 + sin^2y)/(5x^2 + y^2)? Solut

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(x^2 + sin^2y)/(5x^2 + y^2) If (x,y) approaches (0, 0) along the x-axis- that is, as (x, 0), the two functions above become 6(0)(x^3)/(2x^4+ 0)= 0 and x^2(sin^2(0))/(x^2+ 2y^2)= 0 for all x so the limit as (x, 0)-> (0, 0) is 0. But in order that the limit itself exist, we must get the same result no matter which path we use to approach (0, 0). On the line y= x the first function becomes 6x(x^3)/(2x^4+ x^4)= x^4(5x^4)= 1/5 so no matter how close to (0,0) we get on that line, the value of the function is 1/5 and does not approach 0. More generally, the best thing to do is to convert to polar coordinates. That way, a single variable, r, measures the distance from (0,0) no matter what the other, heta, is. (In Cartesian coordinates distance is a combination of x and y.) rac{6yx^3}{2x^4 + y^4}= rac{r sin( heta)r^3cos^3( heta)}{2r^4cos^4( heta)+ r^4 sin^4( heta)} = rac{sin( heta)cos^3( heta)}{2cos^4( heta)+ sin^4( heta)} Notice that the "r" terms have cancelled out- the value, no matter how close to (0,0) we are, depends upon heta. Again, there is no one value we approach as r goes to 0 and so no limit. For the second function, rac{x^2sin(y)}{x^2+ 2y^2}= rac{r^2cos^2( heta)sin(rcos( heta))}{r^2 (cos^2( heta)+ 2sin^2( heta)} = rac{cos( heta)sin(rcos( heta)}{cos^2( heta)+ 2sin^2( heta)}

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