Now this is an example that I don't really understand. It says prove that the limit of (4n^(3)+3n)/(n^(3)-6)(adsbygoogle = window.adsbygoogle || []).push({});

2. Relevant equations

Lim{a}=L

For any epsilion>0 there is a N>0 such that n>N=> |{a}-L|<epsilion

3. The attempt at a solution

So here's the book's solution.

|(4n^(3)+3n)/(n^(3)-6) - 4|=|(3n+4)/(n^(3)-6)|<epslion

but if n>or=2, (3n+4)/(n^(3)-6)<epslion

they found an upper bound for the sequence.

(3n+24)<or =30n

& (n^(3)-6)>or =(1/2)n^(3)

So, (3n+4)/(n^(3)-6)<or= (30n)/( (1/2)n^(3))<epsion

=> 60/n^(2)<epslion

=> (60/epslion)^(-1/2)<n

so this implies that we make N=max{2, (60/epslion)^(-1/2)}

Now here's my solution,

I never found a reason why they used an upper bound. Wouldn't a lower bound work just as well?

(3n+4)/(n^(3)-6)<epslion

n/(n^(2)) <(3n+4)/(n^(3)-6)<epslion. Now I choose the denominator,n^2, arbitaritly just to make just sure that the denominator becomes smaller.

So solving, 1/epslion<n. This implies that we make N=max{2, 1/epslion}

So far, I'm convinced that my solution is the most simple. But is it correct?

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# Homework Help: Proof of a limit.

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