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,

$| a_{n} - L | = ∣ ∣ ∣ \frac{2 sin n}{n} - 0 ∣ ∣ ∣ = ∣ ∣ ∣ \frac{2 sin n}{n} ∣ ∣ ∣$ $abs(a_n-L) = abs((2sinn)/n - 0) = abs((2sinn)/n)$ .

For positive $n$ $n$ , this is $= \frac{2}{n} | sin n |$ $= 2/nabs(sin n)$

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

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

$∣ ∣ ∣ \frac{2 sin n}{n} ∣ ∣ ∣ = \frac{2}{n} | sin n | \leq \frac{2}{n}$ $abs((2sinn)/n) = 2/nabs(sin n) <= 2/n$ .

By making $M$ $M$ sufficiently great, we can make $\frac{2}{n} < ε$ $2/n < epsilon$ for ann $n \geq M$ $n >= M$ .

(Take $M$ $M$ to be any number greater than $\frac{2}{ε}$ $2/epsilon$ .)

Answer 2 · 2017-01-10T00:43:43

Very very simply

Explanation:

We already must know that $sin x$ $sin x$ is between -1 and +1 for every real x, so that $- 1 \leq sin 2 n \leq 1$ $-1 <= sin2n <=1$ for every n (or indeed any real number). When you divide that by n as $n \to \infty$ $n->∞$ , 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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