Given the string an=(n+1)/(5n+2), verify that the limit is 1/5 using epsilon-delta theory.

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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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