> In the current presentation I am working on, quotients > drop out before the limits are taken. So there is no need > to explain the limit of the quotient of two quantities, > both of which are approaching zero at the same time.
You can find "secant --> tangent" arguments in most of the older analytic geometry texts (specifically, U.S. texts roughly dating from 1910 to 1950), where it was considered as not employing calculus methods, even though limits were involved (but only for polynomial functions that can be dealt with algebraically).
Even less calculus needs to be brought onto the scene for sufficiently simple functions. For example, no limits are needed to determine the maximum or minimum of quadratic functions, as this can be done by completing the square and using observations such as: The values of -2(x - 3)^2 + 5 are always less than or equal to 5.
I wrote about a slightly more involved method that works for more complicated functions nearly 4 years ago in an AP-calculus post (archived at Math Forum). Below, between the horizontal ##-lines, is an excerpt from that post. The entire post is at this URL:
One advantage of looking at something like what I give below is that it certainly qualifies (in today's high school environment) as enrichment material, and it also serves to greatly strengthen algebraic skills in a context where you are actually applying those skills.
Mike's function, f(x) = (x-1)(x-2) / (x+1)(x+2), is a quadratic divided by a quadratic. This is a type of function I've seen discussed in a number of papers in old volumes of elementary journals and in old algebra texts. An example of such a paper is Darboux's paper below, but I've seen many others as well. I've also seen quadratics divided by quadratics on a lot of French admissions/scholarship exams that one can find reprinted in "Journal de Mathematics Elementaries", "Mathesis Recueil Mathematique", etc. These papers, texts, and exam appearances are mostly from the 1880s to the early 20th century, by the way.
Those interested in a challenge might want to consider how you can use non-calculus concepts (quadratic formula and max/min of parabolas topics) to investigate where the graph of y = (ax^2 + bx + c)/(dx^2 + ex + f) has a local maximum value and/or a local minimum value. This used to be a standard problem given in "contests for admission" to French universities, starting around 1880 or so through at least (I think) the 1920s. The topic can also be found in most of the more advanced algebra texts from this period, many dozens of which have now been digitized by google.
No one replied that I know of, so I'll demonstrate the non-calculus method using Mike's function.
I'll go through the calculus method first.
[calculus solution omitted in math-teach re-posting]
For the "precalculus method", we note that (for sufficiently nice functions) we can detect the presence of a strict local extrema by investigating the intersection of a variable horizontal line with the graph as the horizontal line is moved vertically. For example, if we get no intersection points with the line y = c when c < 4, exactly one intersection point with the line y = 4, and exactly two intersection points with the line y = c when c > 4, then y = 4 is a local minimum value. The same is true if, instead of "each c < 4" and "each c > 4", this holds for all values of c in some left neighborhood of 4 and all values of c in some right neighborhood of 4.
We begin by rewriting the equation in implicit form as a quadratic expression in the variable x:
x^2y + 3xy + 2y = x^2 - 3x + 2
(y-1)x^2 + 3(y+1)x + 2(x-1) = 0
We can see from this that, for each value of y, there exist at most 2 values of x such that the pair (x,y) belongs to the graph. Since specifying a value of y corresponds to looking at the intersection of the graph with a certain horizontal line, it follows from the previous paragraph that we want to determine for which values of y there is no solution for x, for which value(s) of y there is exactly one solution for x, and for which values of y there are two solutions for x.
But this is exactly what the discriminate of the quadratic tells us!
If b^2 - 4ac < 0, there are no (real) solutions.
If b^2 - 4ac = 0, there is exactly one solution.
If b^2 - 4ac > 0, there are two solutions.
In our case,
b^2 - 4ac = 9(y+1)^2 - 8(y-1)^2
= 9(y^2 + 2y + 1) - 8(y^2 - 2y + 1)
= y^2 + 34y + 1
This is equal to zero when
y = [-34 +/- sqrt(34*34 - 4*1*1)] / 2
y = -17 +/- sqrt(1152)/2
y = -17 +/- 12*sqrt(2),
the same values of y we got earlier.
Also, since b^2 - 4ac = y^2 + 34y + 1 is positive for y < -17 - 12*sqrt(2) (I know this immediately from knowing what the graph of an upward opening parabola looks like relative to the horizontal axis), zero for y = -17 - 12*sqrt(2), and negative for all values of y in a certain right neighborhood of -17 - 12*sqrt(2), I know that y = -17 - 12*sqrt(2) corresponds to a local maximum value.
To find the x-coordinate for this local maximum, we substitute y into the original equation and solve for x.
(y-1)x^2 + 3(y+1)x + 2(x-1) = 0
The task is much easier than you might think, since we already know that the square root part of the quadratic formula is zero.