Proof Limit of (4n^(3)+3n)/(n^(3)-6): Simple Solution

[torquerotates]

Now this is an example that I don't really understand. It says prove that the limit of (4n^(3)+3n)/(n^(3)-6)

Homework Equations

Lim{a}=L

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

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?

 

[konthelion]

$$\epsilon$$ is "epsilon" and you can simply write "<or = " as "<=".

Notice that your sequence is monotonically increasing. Let $$S= \left\{ a_{n}|n \in N \right\}$$ so $$\epsilon > 0,$$ you need to find an index N such that

$$|a_{n}- l| < \epsilon$$

which is the same as

$$a_{n} \leq l < l + \epsilon$$

Notice that l is an upper bound for the set S, so

$$\implies l - \epsilon < a_{n} < l+ \epsilon$$
So, l is the least upper bond or supS . Thus, there is an index N such that $$l - \epsilon < a_{N}$$; but, the sequence is mononotically increasing so

$$l - \epsilon < a_{N} \leq a_{n}$$

In Short: Since your sequence is monotonically increasing, then it converges if and only if it is bounded above.