Prove a limit by the limit theory

[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...

 

[phreak]

So we choose arbitrary \epsilon &gt; 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&gt;N,\, |f(n)-1/2|&lt;\epsilon, that is, |\frac{n^2-1}{2n^2+3}-1/2|&lt;\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} &gt; 1/2-\epsilon. I think you can do the rest. (You also need to prove that 1/2 &gt; \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&gt;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.

 

[kesun]

So we choose arbitrary \epsilon &gt; 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&gt;N,\, |f(n)-1/2|&lt;\epsilon, that is, |\frac{n^2-1}{2n^2+3}-1/2|&lt;\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} &gt; 1/2-\epsilon. I think you can do the rest. (You also need to prove that 1/2 &gt; \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&gt;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.

I get what you have said and proved that 1/2 &gt; \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.. :)