Proving a sequence converges using epsilon-N definition.

Question

I'm stuck with what to do next in my homework problem please help.

Answer 1

$\textbf{Hint:}$Note that for big $j$ you have:

$$4-\frac{2}{j}+\frac{20}{j^2}>3$$

(you can check this) and:

$$1-\frac{6}{j}<\frac{3}{2}$$

So:

$$\left|\frac{1-\frac{6}{j}}{4-\frac{2}{j}+\frac{20}{j^2}}\right|<\frac{\frac{3}{2}}{3}=\frac{1}{2}$$

So for big $j$'s:

$$\frac{1}{j}\left|\frac{1-\frac{6}{j}}{4-\frac{2}{j}+\frac{20}{j^2}}\right|<\frac{1}{j}\frac{1}{2}$$

Answer 2

Rewrite $$ \frac{j^2+2}{2j^2-j+10} = \frac{1}{2} + \frac{{j\over 2} - 3}{2j^2-j+10} $$. To find an $N(\epsilon)$ that works to prove the limit, a nice trick is to solve for $N$ the equation $$ \frac{N^2+2}{2N^2-N+10} = \epsilon $$ IN the course of solving this, you can always discard terms that make $N$ larger. You end up with something like $$N(\epsilon) = \frac{1}{8\epsilon} + \frac{1}{2} $$

Then you plug that back in and demonstrate that the distance is less than $\epsilon$.