Find N(a) such that f(n)<a for all n>N(a) where f(n)=1/(5n^2-8).

Question

n are natural numbers, and a is a real positive number.

I can find N in terms of a but this question is from a epsilon-delta limit chapter, so I have no idea how to find it whilst relating it to this topic.

Answer 1

Given: f(n) = 1/(5n² - 8)

We need to show that, as n → ∞, f(n) → 0

Formally, this means, given any number a, there is some number, N(a), such that,∀ n > N(a), f(n) < a

Now, f(n) < a iff 1/(5n² - 8) < a

iff 5n² - 8) > 1/a

iff n² > (1/a + 8) / 5 = 1/5a + 8/5

iff n > +√(1/5a + 8/5)

Choose N(a) = [+√(1/5a + 8/5)] + 1 where [ ] is the floor function, i.e., the greatest integer that is not > a

Then n > N(a) ⇒ f(n) < a