20分的ε-N definition

Notice that |(3n^2 + 2) / (4n^2 + 3) - 3/4| = | 1 / (16n^2 + 12)| < | 1 / (16n^2)|
So we could take N = upperint [1 / 4 sqrt(k) ]
(Note: sqrt(k) = root k, and upperint(x) = the least integer greater than x)

so for any ε > 0, n >= N implies
|(3n^2 + 2) / (4n^2 + 3) - 3/4| = | 1 / (16n^2 + 12)|
< | 1 / (16n^2)|
< ε

So by definition we have
lim (3n^2 + 2) / (4n^2 + 3) = 3/4
n->∞

Use ε-N definition to prove that
lim (3n^2 + 2) / (4n^2 + 3) = 3/4
n → ∞ .

(3n^2 + 2) / (4n^2 + 3)
= (3n^2 + (9/4) – (1/4) ) / (4n^2 + 3)
= (3/4) – { 1/ 4(4n^2 + 3) }
Now, given any ε>0 , take N > √{ (1/4ε) -3} /2 , then |(3n^2 + 2) / (4n^2 + 3) - 3/4| < | 1/ 4(4n^2 + 3) | < |1/ 4(4N^2 + 3) | < ε whenever n > N
圖片參考：http://i175.photobucket.com/albums/w130/bjoechan2003/My%20Cat/DSC00020-1.jpg?t=1169897591

lim (3n^2 + 2) / (4n^2 + 3)
n-&gt;∞
= lim n^2(3 + 2/n^2) / n^2(4 + 3/n^2)
n-&gt;∞
= lim (3 + 2/n^2) / (4 + 3/n^2) [eliminating n^2]
n-&gt;∞
= 3 / 4 [because lim 2/n^2 = 0 and lim 3/n^3 =0
.........................n-&gt;∞..................n-&gt;∞

Take N &gt; (1 - 12ε)/(16ε)
You will have |(3n^2 + 2) / (4n^2 + 3) - 3/4 | &lt; ε