How do you prove that the limit of #(x^2)sin(1/x)=0 # as x approaches 0 using the epsilon delta proof?
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To prove that the limit of (x^2)sin(1/x) as x approaches 0 is 0 using the epsilon-delta proof, we need to show that for any given epsilon greater than 0, there exists a delta greater than 0 such that if 0 < |x - 0| < delta, then |(x^2)sin(1/x) - 0| < epsilon.
Let's proceed with the proof:
Given epsilon > 0, we need to find a delta > 0 such that if 0 < |x - 0| < delta, then |(x^2)sin(1/x) - 0| < epsilon.
First, note that |sin(1/x)| ≤ 1 for all x ≠ 0.
Now, let's consider |(x^2)sin(1/x) - 0| = |x^2sin(1/x)| = |x^2||sin(1/x)|.
Since |sin(1/x)| ≤ 1, we have |x^2||sin(1/x)| ≤ |x^2|.
To ensure that |x^2| < epsilon, we can choose delta = sqrt(epsilon).
Now, if 0 < |x - 0| < delta = sqrt(epsilon), then |x^2| < epsilon.
Therefore, we have shown that for any given epsilon > 0, there exists a delta > 0 (specifically, delta = sqrt(epsilon)) such that if 0 < |x - 0| < delta, then |(x^2)sin(1/x) - 0| < epsilon.
Hence, the limit of (x^2)sin(1/x) as x approaches 0 is indeed 0, as proven using the epsilon-delta proof.
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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.
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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