# How do you prove by definition that the function #f(x)= x^2 sin (1/x)# is continuous at x=0?

Not continuous

This function would be continuous for example,

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If you re-define the function as Jim suggested, then it would be continuous at

Method 2 (use the Squeeze Theorem ):

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To prove that the function f(x) = x^2 sin(1/x) is continuous at x = 0, we need to show that the limit of f(x) as x approaches 0 exists and is equal to f(0).

First, let's evaluate the limit of f(x) as x approaches 0. We can rewrite the function as f(x) = x^2 * sin(1/x) = x * (x * sin(1/x)).

Now, we know that the limit of sin(1/x) as x approaches 0 is 0. This means that the term (x * sin(1/x)) approaches 0 as x approaches 0.

Next, we multiply this term by x, which also approaches 0 as x approaches 0. Therefore, the entire expression x * (x * sin(1/x)) approaches 0 as x approaches 0.

Since the limit of f(x) as x approaches 0 is 0, we can conclude that f(x) is continuous at x = 0.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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