imagine you've got the function x plotted in a graph(you can actually draw it in a piece of paper). and you want to show that <br />
\lim_{x \to 1} x \neq 2. First, set an interval ]1-delta,1+delta[ in x axes, see the image of the 1+delta; epsilon will be=|f(1+delta)-f(1)|. (in general, it will be the max{|f(1+delta)-f(1)|,|f(1-delta)-f(1)|}, in this case, the function is similar to the right and left of f(1), so, only one part is needed because both are equal).
Now, you see that if the delta is very big, say 4, then epsilon=4 rigth?! so, (the second part of the condition, the epsilon part) is |2-f(1)|=2-1=1<epsilon: the condition is verified.(2 because you are testing the condition on the point 2 you are asking: "is it really true that the condition is verified to EVERY delta?")
Note however that if you put delta lower, say 0.5, epsilon is 0.5 and |2-f(1)|=|2-1|>epsilon:. Exists a delta that don't verify the condition, that implies that isn't true that for every delta, exists..bla bla bla...so, the limit isn't 2.
Hope this helps solve your problem...
and hope i didn't make any mistake...^_^