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We'll have to show that for any positive number epsilon, there is N = N(epsilon), so that:
|an - 1/5| < epsilon for any n > N(epsilon)
We'll substitute an by it's given expression:
|(n+1)/(5n+2) - 1/5| = |(n+1-5n-2)/5(5n+2)| = 3/5(5n+2)
We'll get the inequality:
3/5(5n+2) < epsilon
3 < 5 epsilon (5n + 2)
We'll remove the bracktes:
3 < 25*epsilon*n + 10 epsilon
We'll subtract 10 epsilon:
25*epsilon*n > 3 - 10 epsilon
n > (3 - 10 epsilon)/25 epsilon
N(epsilon) = 3/25epsilon - 2/5
For all terms that have n > N(epsilon) = 3/25epsilon - 2/5,
|an - 1/5| < epsilon
We say that Lt x-->n f(x) = l, if for any given e however small, we can always find |x-a| < d(delta).
In the given case Lt n--> infinity an = (n+1)/(5n+2) = Lt (1/n) -> 0 (1+1/n)/(5+2/n)
= lt x-->0 (1+x)/(5+2x) = 1/5. Let us take |an-0| < e = 0.01.
Let is take |(1+x)/(5+2x) - 1/5| < 0.01
=> |(5+5x-5-2x)| < (25+10x) 0.01
=> |3x| < 0.25+x
=> 3x-0.1x < 0.25
=> 2.9x < 0.25.
=> x < 0.25/2.90 < 0.09.
Therefore we are able to find a d such that x-0< 0.09, Lt x-> 0 (1+1)/(5+2x) = 1/5.
Similarly for any e given an-1/5 < e we can find a positive d such that |x-0| < d.
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