Question #57453

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Question #57453

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Answer 1 · 2017-01-09T23:38:13

See explanation section below.

Explanation:

In this case,

$abs(a_n-L) = abs((2sinn)/n - 0) = abs((2sinn)/n)$ .

For positive $n$ , this is $= 2/nabs(sin n)$

Recall that for all real $n$ , we have $abs(sin n) <= 1$

So for all $n > 0$ , we know that

$abs((2sinn)/n) = 2/nabs(sin n) <= 2/n$ .

By making $M$ sufficiently great, we can make $2/n < epsilon$ for ann $n >= M$ .

(Take $M$ to be any number greater than $2/epsilon$ .)

Answer 2 · 2017-01-10T00:43:43

Very very simply

Explanation:

We already must know that $sin x$ is between -1 and +1 for every real x, so that $-1 <= sin2n <=1$ for every n (or indeed any real number). When you divide that by n as $n->∞$ , then it is clear that the answer is zero. "Everything should be made as simple as possible, although not any simpler"

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Question #57453

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