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Let \(f: R^{2} \rightarrow R\) be given by $$ f(x, y)=x^{2}+y^{2} $$ Show that \(\mathrm{f}\) is continuous at \((0,0)\).

Short Answer

Expert verified
Using the limit definition of continuity, we demonstrated that the value of the function \(f(x, y) = x^2 + y^2\) at \((0,0)\) is equal to the limit of the function as \((x, y)\) approaches \((0,0)\). By converting to polar coordinates and finding the limit, we showed that \(f(0,0) = \lim_{(x, y) \to (0,0)} f(x, y) = 0\), thus proving \(f(x, y)\) is continuous at \((0,0)\).
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Step 1: Find the value of the function at the point \((0,0)\)

To find the value of \(f(x, y)\) at the point \((0,0)\), substitute \(x = 0\) and \(y = 0\) into the expression for \(f(x, y)\): $$ f(0, 0) = (0)^2 + (0)^2 = 0 $$

Step 2: Determine the limit of the function as \((x, y)\) approaches \((0,0)\)

We must show that the limit of \(f(x, y)\) as \((x, y)\) approaches \((0,0)\) exists and equals the value of the function at the point \((0,0)\). Using polar coordinates, let \(r = \sqrt{x^2 + y^2}\) and \(\theta\) be any angle. Then, the limit can be expressed as: $$ \lim_{(x, y) \to (0,0)} f(x, y) = \lim_{r \to 0} r^2 $$

Step 3: Evaluating the limit

To find the limit, we simply substitute \(r = 0\): $$ \lim_{r \to 0} r^2 = 0^2 = 0 $$

Step 4: Comparing the function value and the limit

Now that we have found both the value of the function at the point \((0,0)\), which is \(f(0,0)=0\), and the limit as \((x, y)\) approaches \((0,0)\), which is \(\lim_{(x, y) \to (0,0)} f(x, y) = 0\), we can conclude that: $$ f(0,0) = \lim_{(x, y) \to (0,0)} f(x, y) $$ Since the value of \(f(x, y)\) at \((0,0)\) is equal to the limit of \(f(x, y)\) as \((x, y)\) approaches \((0,0)\), the function is continuous at \((0,0)\).

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