# Prove a limit by the limit theory

1. Mar 5, 2009

### kesun

Apply the limit definition to prove $$lim_{n\rightarrow\infty}\frac{n^{2}-1}{2n^{2}+3}=\frac{1}{2}$$

(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 $$\epsilon$$ then somehow prove that |f(n) - 1/2| < $$\epsilon$$. I need to know what are the exact steps to prove stuff like this...

2. Mar 5, 2009

### phreak

So we choose arbitrary $$\epsilon > 0$$. This means that $$\epsilon$$ can be ANYTHING positive. Now, given this $$\epsilon$$, we want to find an $$N \in \mathbb{N}$$ such that for all $$n>N,\, |f(n)-1/2|<\epsilon$$, that is, $$|\frac{n^2-1}{2n^2+3}-1/2|<\epsilon.$$ Now, fortunately for us, our function is monotone increasing, so all we have to find is a number N satisfying $$\frac{N^2-1}{2N^2+3} > 1/2-\epsilon$$. I think you can do the rest. (You also need to prove that $$1/2 > \frac{n^2-1}{2n^2+3}$$ for all n for this proof to work.)

The concept of the proof is as follows. Since $$\epsilon>0$$ 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 $$\epsilon$$. This just says that as N gets increasingly higher, the value of the function gets increasingly close to 1/2.

Last edited: Mar 5, 2009
3. Mar 5, 2009

### kesun

I get what you have said and proved that $$1/2 > \frac{n^2-1}{2n^2+3}$$ 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.. :)