How do I use polar coordinates to prove that the limit of the function f(x,y)=(2xy)/(x^2 +y^2) doesn't exist as it approaches (0,0)?
well, we know that since xy > 0 if x and y have the same sign, and xy < 0 otherwise, it is clear that the limit does not exist.
since 2xy = sin2θ, the same applies. sin2θ/r^2 can be either positive or negative, depending on θ, so the limit does not exist.
How did you get 2xy=sin2θ? Should 2xy = 2r^2 cosθsinθ?
oh, yeah. You are correct. 2xy = r^2 sin2θ
so, f(x,y) = sin2θ
To prove that the limit of the function \(f(x,y) = \frac{{2xy}}{{x^2 + y^2}}\) doesn't exist as it approaches \((0,0)\) using polar coordinates, follow these steps:
1. Convert the Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\). Recall that \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\).
2. Rewrite the function in terms of \(r\) and \(\theta\). Substitute \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\) into the function \(f(x,y)\).
\(f(r, \theta) = \frac{{2(r\cos(\theta))(r\sin(\theta))}}{{(r\cos(\theta))^2 + (r\sin(\theta))^2}}\)
Simplify this expression by canceling out the common factor of \(r\).
\(f(r, \theta) = \frac{{2r^2\cos(\theta)\sin(\theta)}}{{r^2(\cos^2(\theta) + \sin^2(\theta))}}\)
Notice that \(r\) cancels out, so the expression becomes:
\(f(r, \theta) = \frac{{2\cos(\theta)\sin(\theta)}}{{\cos^2(\theta) + \sin^2(\theta)}}\)
Further simplifying, we see that \(\cos^2(\theta) + \sin^2(\theta) = 1\):
\(f(r, \theta) = 2\cos(\theta)\sin(\theta)\)
3. Now, to show that the limit does not exist as \((r, \theta)\) approaches \((0, 0)\), we need to consider different paths in the polar coordinate system.
a) Approach along the \(x\)-axis (\(\theta = 0\)):
In this case, \(\cos(\theta) = 1\) and \(\sin(\theta) = 0\).
\(f(r, \theta) = 2\cos(\theta)\sin(\theta) = 2(1)(0) = 0\).
b) Approach along the \(y\)-axis (\(\theta = \frac{\pi}{2}\)):
In this case, \(\cos(\theta) = 0\) and \(\sin(\theta) = 1\).
\(f(r, \theta) = 2\cos(\theta)\sin(\theta) = 2(0)(1) = 0\).
c) Approach along the line \(y = x\) (\(\theta = \frac{\pi}{4}\)):
In this case, \(\cos(\theta) = \sin(\theta) = \frac{\sqrt{2}}{2}\).
\(f(r, \theta) = 2\cos(\theta)\sin(\theta) = 2\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) = 1\).
As we can see, the value of the function \(f(r, \theta)\) depends on the chosen path, which means the limit as \((r, \theta)\) approaches \((0, 0)\) does not exist.
By considering different values of \(\theta\) (representing different paths), we can see that the function value varies and does not approach a specific value, thereby proving that the limit of the function \(f(x,y) = \frac{{2xy}}{{x^2 + y^2}}\) does not exist as \((x, y)\) approaches \((0,0)\).