# Proving a sequence has limit using an epsilon - N argument

I have two separate sequences that I using this approach on. $$a_n=\frac{n^3-2n^2+3}{2n^3+7n}$$ & $$a_n=\frac{n^3}{2^n}$$

Proof 1

$$a_n=\frac{n^3-2n^2+3}{2n^3+7n}\to \frac{1}{2}$$

$$\text{Let } \epsilon \gt 0 \,\, \exists N \;\;\forall n\geq N \; \left\lvert {a_n-\frac{1}{2}}\right\rvert \lt \epsilon$$ $$\left\lvert {\frac{n^3-2n^2+3}{2n^3+7n}-\frac{1}{2}}\right\rvert=\left\lvert {\frac{(2n^3-4n^2+6)-(2n^3+7n)}{4n^3+14n}}\right\rvert=\left\lvert {\frac{-4n^2-7n+6}{4n^3+14n}}\right\rvert \lt \frac{1}{2n} \lt \epsilon$$ $$\frac{1}{2\epsilon} \lt n$$ I am having difficulty in showing exactly why, $$\left\lvert {\frac{-4n^2-7n+6}{4n^3+14n}}\right\rvert \lt \frac{1}{2n}$$

Proof 2

$$a_n=\frac{n^3}{2^n} \to 0$$ $$\text{Let } \epsilon \gt 0 \,\, \exists N \;\;\forall n\geq N \; \left\lvert {a_n-0}\right\rvert \lt \epsilon$$ $$\left\lvert {\frac{n^3}{2^n}}\right\rvert=\frac{n^3}{2^n} \lt \frac{?}{?} \lt \epsilon$$ I am lost as to where to proceed on this one. Any input would be greatly appreciated . Thanks

• The numerator in part 1 should be $-4n^2-1$. Mar 29 '19 at 23:27
• Just edited the sequence. I had copied it wrong here Thank you for catching that
– MAC
Mar 29 '19 at 23:34
• I wasn't talking about the sequence, but about the algebra mistake you made in the numerator in the first exercise. Mar 30 '19 at 0:07
• @JohnWaylandBales the algebra mistake is result of the sequence mistake.
– MAC
Mar 30 '19 at 2:21

It would be simplest to show that $$\left|\dfrac{4n^2+1}{4n^3+14n}\right|<\dfrac{1}{n}$$ since, clearly $$4n^3+n<4n^3+14n$$.
For $$n\ge16$$ we have that $$n^4\le2^n$$ from which it follows that $$\dfrac{n^3}{2^n}\le\dfrac{1}{n}$$.
• Divide both side of the inequality $4n^3+n<4n^3+14n$ by $n(4n^3+14n)$. Mar 30 '19 at 5:01