Question #57453

2 Answers
Jim H
Jan 9, 2017

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#.)

Reg H.
Jan 10, 2017

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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