I have a quotient:

(2n^4 + 4n^2 + 3n - 5)/(n^4 - n^3 + 2n^2 - 80)

To show that it converges to 2 by "first principles" I cannot simply prove by the algebra of limits can I? That is, to divide throughout by the highest power of n (the most influential) and then prove each remaining term's (either 1/n^x -> 0 or c -> c) limit. By doing it this way It is rather easy to show that it converges to 2.

However, I have to admit, there must be a harder way!

I was thinking along the lines of exactly how the algebra of limits is proven itself! For example (as I noted above) the limit of a constant is a constant. How is this proven?

Fix e>0, we want an N E N such that n >_ N

Hence |c - c| < e

<=> 0 < e therefore any N will do and so I have shown that the limit of a constant is itself.


Furthermore, I can show rigourously that the limit of 1/n^r is zero.

So would I have to incorporate each individual proof of each limit of each term in the quotient to show by first principles that the quotient converges to zero?

I would I simply treat the quotient as a whole and find another quotient which is always bigger than (but much simpler than) the original. And then apply the Order Axioms to epsilon?

Or is there an easier way?

By first principles they usually mean you want to show directly that your function satisfies the definition of the limit, so if I give you an e>0, you need to produce an N that will give e>|f(n)-2|for all n>=N.

Just look at |f(n)-2|. Show that you can bound it by something simpler that goes to zero. Figure out the N that will work for this new function. Translate this back to the N you're looking for with |f(n)-2|.

I don't know if this helps(as I've forgotten how what first principles means), but my approach to this would be as follows

2n^4+4n^2+3n+5/n^4-n^3+2n^2-80=2 +(2n^3+3n+155/n^4-n^3+2n^2-80)

so the limit = 2 + lim(2n^3+3n+155/n^4-n^3+2n^2-80)

This smaller limit is much easier to work with.

Cheers,

Fix e > 0. I have to produce an N that will give |f(n) - 2| < e

|f(n) - 2| = (2n^3 + 3n + 155)/(n^4 - n^3 - 2n^2 - 80) you can check this but I think it is right.

This function is always positive for any n > 3 so I can take away the absolute values. To give...

(2n^3 + 3n + 155)/(n^4 - n^3 - 2n^2 - 80) < e

As long as n > 3 => n^4 - n^3 - 2n^2 - 80 >_ 0, the following inequality holds...

(2n^3 + 3n + 155)/(n^4 - n^3 - 2n^2 - 80) < 2n^3 / n^4 < e

Thus, the function is bounded above by 2/n which converges to zero

Hence e > 2/n

<=> n > 2/e

Now take N to be any integer greater than n; therefore N is sufficient to show that n^3/n^4 < e. Since N works for this it should work for (2n^3 + 3n + 155)/(n^4 - n^3 - 2n^2 - 80) for any n > 4.

How does this look???

As long as n > 3 => n^4 - n^3 - 2n^2 - 80 >_ 0, the following inequality holds...

(2n^3 + 3n + 155)/(n^4 - n^3 - 2n^2 - 80) < 2n^3 / n^4 < e


This bolded inequality doesn't hold. Put in any value of n>3 and you'll see the inequality is actually the other way around. (2n^3 + 3n + 155)>2n^3 and n^4>(n^4 - n^3 - 2n^2 - 80). So you've made the numerator smaller and the denominator larger, making the entire quotient smaller.

Try to bound the numerator from above by a multiple of n^3 and the denominator from below by a multiple of n^4.

If it was me, I'd just use your first mentioned easy method.

But maybe I'm just lazy :)