Find the limit of An=(n^2-3n+5)/(3n^2+n-1) as n goes to infinity. We know the li

Question

Find the limit of An=(n^2-3n+5)/(3n^2+n-1) as n goes to infinity.

We know the limit (An)=1/3 if we divide An by n^2. Now to prove it we know there exists an epsilon>0 for all N such that
|An-1/3|
So then |[(n^2-3n+5)/(3n^2+n-1)]-1/3| |[3n^2-9n+15-(3n^2+n-1)]/(9n^2+3n-3)|=
|(-10n+16)/(9n^2+3n-3)|< # ? <epsilon

Here is where I am stuck because I don't know what to choose to be less than this function. If I choose 1/(9n^2) this is bigger not less so please explain??? How to prove it converges to 1/3 and how to choose this # and why??? Thank you

Explanation / Answer

|(-10n+16)/(9n^2+3n-3)|=|10n-16|/(9n^2+3n-3)<10n/(9n^2+3n-3)

The idea is to bound the fraction using the denominator

note that 9n^2+3n-3>n^2 or

8n^2+3n-3>0 is true for any n>0

Hence

10n/(9n^2+3n-3)<10n/n^2=10/n

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