In high school students are taught that division by 0 is undefined. Why would division by 0 be considered undefined? The answer, in part, lies in the definition of division. When division is first taught, a teacher may take 6 apples and arrange them groups of 2 apples each. The teacher then points out that there are 3 groups of apples. What we learn from these examples is that we can add the number 2 three times to obtain 6, so 6/2 = 3. In mathematical notation, a/b = c if b*c=a. Now what happens when we divide by 0? If we have 6 apples and we divide them into groups of 0, how many groups can we make before our sum is greater than 6? First we take 0 apples and put them in a group. We still have six apples left, so we take another group of 0 apples. We still have 6 apples left. Perhaps you are seeing the problem. There is no number c such that c*0 = a, thus we are told that division by 0 is undefined.
Or is it? Consider the limit as x approaches 0 of 1/x. This is well known to be infinity. Also, the limit is approaching 1/0. Thus in calculus we can define division by zero to be infinity. Now there are some who would quibble over whether the infinity is positive or negative, but as any geometer knows there is really no difference.
The next question we must ask is, "Is this a good definition for division by 0?" Well, what is a "good" definition. Generally, we consider a definition good if it works in the problems we want to solve. The natural log is defined as an integral, because that makes it the inverse function of e^x: it works. This definition for division by 0 happens to be a good definition (at least for a large number of problems).
This leads to an interesting question: are there operations for which good definitions do not exist? The answer is yes. There are seven problems known as indeterminate forms for which good definitions do not exist. The forms are listed below.
Why are these forms such a problem? Lets consider them one at a time.
What is the problem with 0/0? Remember the definition of division: a/b=c if c*b=a. In this case 0/0=c iff c0=0. The problem is that there is not a unique answer c. c=0, c=1, c=3.141592654, and any other real number work. Thus the answer is not unique which is, in simple terms, bad.
The inf/inf, 0*inf forms are just rearranged versions of 0/0 and thus suffer from the same problem. To see this note that inf/inf = (1/0)/(1/0) = 0/0. Also 0*inf = 0*(1/0) = 0/0.
The problem with 0^0 is a conflict of definitions. 0^k=0 for all real k. k^0=1 for all real k. With 0^0 which definition do we use? Remember also that a^b can be defined over all real numbers a and b by a^b=e^(b*ln(a)). Since ln(0) is undefined we have a problem right here in this town (whatever town this may be).
inf^0 suffers the same problem as 0^0. inf^0 = (1/0)^0 = (1^0)/(0^0).
1^inf is a little bit more interesting. 1^k=1 for all real numbers k. However lim as k->inf of (1+1/k)^k = 1^inf = e. This is one of the well-known sequence approximations for e. Once again we have two definitions which conflict: bad.
The final form, inf-inf, runs contrary to a property of real numbers learned in elementary school. First suppose inf-inf=0; this would seem reasonable since a-a = 0 for all real numbers. Then inf-inf+inf = (inf-inf) + inf = 0+inf = inf. But inf-inf+inf = inf - (inf + inf) = inf-inf = 0. Whooops.
Now none of these dueling definitions would be cause for concern if they never appeared in problems, but they do. It is for this reason that L'Hopital's rule is so important to the students of calculus.