# Using the definition of convergence, how do you prove that the sequence #limit (sin n)/ (n) = 0# converges from n=1 to infinity?

To prove that the sequence limit (sin n)/n = 0 converges from n = 1 to infinity using the definition of convergence, we need to show that for any ε > 0, there exists an N such that for all n ≥ N, |(sin n)/n - 0| < ε.

Given ε > 0, we aim to find N such that |(sin n)/n - 0| < ε for all n ≥ N.

Notice that |(sin n)/n| ≤ 1/n for all n ≥ 1 (as |sin n| ≤ 1 for all n).

Let's choose N = 1/ε. Then for all n ≥ N,

|(sin n)/n - 0| = |(sin n)/n| ≤ 1/n < 1/(1/ε) = ε.

Therefore, for any ε > 0, we have found an N (specifically N = 1/ε) such that for all n ≥ N, |(sin n)/n - 0| < ε.

Hence, by the definition of convergence, the sequence limit (sin n)/n = 0 converges from n = 1 to infinity.

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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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