How Can I Prove the Convergence of a Fraction with Large Exponents?

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SUMMARY

The limit of the expression (2n^4 + 4n^2 + 3n - 5)/(n^4 - n^3 + 2n^2 - 80) converges to 2 as n approaches infinity. This conclusion is derived using the algebra of limits, where the numerator simplifies to 2 and the denominator simplifies to 1. A more rigorous approach involves fixing epsilon (e) > 0 and analyzing the absolute value of the quotient minus 2, although this is not strictly necessary for proving convergence. The discussion emphasizes the importance of understanding the behavior of terms as n increases, particularly the diminishing influence of 1/n^r terms.

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I need some help with a question.

Q) Prove that (2n^4 + 4n^2 + 3n - 5)/(n^4 - n^3 + 2n^2 - 80) converges to 2 as n goes to infinity.

A)

By the algebra of limits, this converges to 2 since

lim(n->oo)[2 + 4/n^2 + 3/n^3 - 5/n^4]/lim(n->oo)[1 - 1/n + 2/n^2 - 80/n^4)

(2 + 0 + 0 + 0)/(1 - 0 + 0 + 0) = 2

However, I would like to do this a little more precisely and rigorously. Can someone tell me...

Would I fix epsilon (e) > 0.

Then the absolute value of the quotient minus 2 must be bigger than epsilon, etc...

Or would I approach this in another way.

Any Help would be appreciated.

Thanks.
 
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Then the absolute value of the quotient minus 2 must be bigger than epsilon, etc...

I think you want smaller, not bigger.
 
You don't want to do that by epsilons not all at once: it will not be pretty or interesting or useful.

you can do something a little more rigorous using the expression after you've divided through by n^4, but it isn't necessary. you can show the top tends to 2 the bottom to 1, and each of those is because the 1/n^r terms tend to zero.The whole point of learning difficult mathematics is to not have to evaluate these kinds of things all at once. It won't help you at all to do so.
 

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