Unfortuately, many beginning students, seeing examples like that, get the impression that "limit" is just a fancy way of talking about the value of the function! That is completely untrue- in fact, in general the limit of function, at, say, x= a, has no relationship to the value there.
The confusion comes because functions for which the limit is the value of the function, called "continuous" functions, are so nice and simple, that we tend to use them all the time! In fact, "almost all" functions are not continuous but almost all the functions we use are continuous.
janeba, this is a very large question (much like your previous question about derivatives!). The "limit" concept really talks about what happens close to a point but there are a few basic properties you can use:
1) If f is a constant function, f(x)= C, then \lim_{x\rightarrow a}f(x)= C for all a.
2) If f(x)= x, then \lim_{x\rightarrow a} f(x)= a.
Those are called the "trivial limits". Notice that if f(x)= C, then f(a)= C and if f(x)= x, then f(a)= a so those functions are "continuous" at all a.
3) If \lim_{x\rightarrow a}f(x)= L and \lim_{x\rightarrow a}g(x)= K then \lim_{x\rightarrow a}(f(x)+ g(x))= L+ K.
4) If \lim_{x\rightarrow a}f(x)= L and \lim_{x\rightarrow a}g(x)= K then \lim_{x\rightarrow a}(f(x)*g(x))= L* K.
5) If \lim_{x\rightarrow a}f(x)= L and \lim_{x\rightarrow a}g(x)= K and K is not 0, then \lim_{x\rightarrow a}(f(x)/g(x))= L/K.
Notice that the last does not help you if the limit in the denominator is 0! Yet, the formula for the derivative always involves a fraction with the denominator going to 0! The last property, and I think the most important, although it is seldom emphasized (I remember a text titled "Calculus for Economics and Business Administration" that did not even mention it!) is
6) If f(x)= g(x) for all x except x= a, then \lim_{x\rightarrow a}f(x)= \lim_{x\rightarrow a} g(x).
You should be able to see how the first 5 of those give the result for polynomials that sutupidmath mentioned- that all polynomials are continuous.
But suppose we were given f(x)= (x^2- 4)/(x-2), f(2)= 100, and asked to find the limit as x goes to 2. (Notice that (2^2- 4)/(2- 2)= 0/0 which is "indeterminant". That's why I gave the separate definition of f(2).) That's a fraction so we would want to use (5) above but we can't. The limit in the denominator is just (by (1) above) 2- 2= 0. What we can do is say that x^2- 4= (x- 2)(x+ 2) and as long as x is not 2[/itex] we can cancel the (x-2) in the numerator with the (x-2) in the denominator as say (x^2-4)/(x-2)= (x-2)(x+2)/(x-2)= (x+2) as long as x is not 2.
When x= 2, f(x)= 100 which is certainly not equal to 2+ 2= 4. But that doesn't matter, since f(x)= x+ 2 for all x except 2, \lim_{x\rightarrow 2} f(x)= lim_{x\rightarrow 2} x+ 2 and the limit on the right is just 2+ 2= 4 so lim_{x\rightarrow 2} f(x)= 4 even though f(2)= 100.
Of course, if I were to take x= 1.9999 or x= 2.00001, since those are not equal to 2 I would use the first formula and I would get (1.9999^2- 4)/(1.9999- 2)= -.00039999/-.0001= 3.9999 and (2.00001^2- 4)/(2.00001- 2)= 0.0000400001/.00001= 4.00001, both very close to 4. The limit tells me what a function is like <b>close to</b> but not at a particular value of x.<br />
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(Thanks for not posting in all capitals!)
