Apply the limit definition to prove [tex]lim_{n\rightarrow\infty}\frac{n^{2}-1}{2n^{2}+3}=\frac{1}{2}[/tex] (question stated above) I started by writing it as |f(n) - 1/2| and attempted to reduce it, but I don't think it's reducible so I am not able to simplify it.. By looking at it further, it stuck me because I don't know where to go with this exactly. I know I am supposed to come up with this arbitrary [tex]\epsilon[/tex] then somehow prove that |f(n) - 1/2| < [tex]\epsilon[/tex]. I need to know what are the exact steps to prove stuff like this...
So we choose arbitrary [tex]\epsilon > 0[/tex]. This means that [tex]\epsilon[/tex] can be ANYTHING positive. Now, given this [tex]\epsilon[/tex], we want to find an [tex]N \in \mathbb{N}[/tex] such that for all [tex]n>N,\, |f(n)-1/2|<\epsilon[/tex], that is, [tex]|\frac{n^2-1}{2n^2+3}-1/2|<\epsilon.[/tex] Now, fortunately for us, our function is monotone increasing, so all we have to find is a number N satisfying [tex]\frac{N^2-1}{2N^2+3} > 1/2-\epsilon[/tex]. I think you can do the rest. (You also need to prove that [tex]1/2 > \frac{n^2-1}{2n^2+3}[/tex] for all n for this proof to work.) The concept of the proof is as follows. Since [tex]\epsilon>0[/tex] can be made as small as we want, for ANY given small number, we can find a number N so high that for every number greater than N, the difference between 1/2 and the value of the function is smaller than the small number [tex]\epsilon[/tex]. This just says that as N gets increasingly higher, the value of the function gets increasingly close to 1/2.
I get what you have said and proved that [tex]1/2 > \frac{n^2-1}{2n^2+3}[/tex] for all n, but I am still not very sure when you said to find an N, do you mean any arbitrary N or is it found by some sophistic method? I am not very getting used to this method of proving limit yet, so I might need a bit more details and a few key steps.. :)