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Problem 57

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

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 $$(0,0)$$.
See the step by step solution

## 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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