the limit as x tends to a of f(x) equals L if;
for every number epsilon (usually, can't find the symbol) > 0, there is a corresponding number [sig] > 0, such that for all x;
0 < |x-a| < [sig] ==> |f(x) - L| < epsilon.
Is the formal definition.
for example (easy one);
To show that the limit of 5x - 3 as x tends to 1 is actually 2;
so a = 1, L = 2 (since this is what it does appear to converge to). Need
0 < |x-1| < [sig] for any epsilon > 0.
f(x) is within epsilon of L ie. |f(x) - 2| < epsilon. So to find [sig] from this,
|(5x-3) - 2| = |5x - 5| < epsilon
5|x - 1| < epsilon
|x - 1| < epsilon / 5.
so [sig] = epsilon / 5
and from 0 < |x - 1| < [sig] = epsilon / 5,
|(5x - 3) - 2| = 5 |x - 1| < 5 (epsilon / 5) = epsilon.
Which proves that L = 2.
Alternatively find a good textbook
