## Why Mathematics Needs Limits

In algebra, sometimes we cannot simply plug a value into a formula and calculate the answer. A common example occurs if we need to “divide by zero”. Remember, if we say “a=b/c”, it means that “b=a*c”. But if ‘c’ has the value of zero, and ‘b’ is non-zero, then there is no number ‘a’ to satisfy “b=a*c”. We then say that “a=b/c” is undefined when ‘c’ has the value zero.

In some situations we can sneak up on a useful answer, however. One method is to use “limits”.

## Karl Weierstrass Set the Limits

Although Newton, Leibniz and Cauchy may be more famous for their work with infinitesimally-small numbers and, therefore, for the mathematics of calculus, it was Weierstrass who generally gets the credit for “epsilon-delta” limits.

Karl Weierstrass (1815-1897) had an interesting life. His father sent him to the University of Bonn to study finance and law. Weierstrass instead engaged in fencing and drinking … presumably in that order, since he survived his sword-play. Later he officially switched to his preferred subject, mathematics, and became a teacher. The University of Königsberg rewarded his paper on Abelian functions with an honorary doctorate in 1854. He soon stopped “teaching” but did lecture as part of his duties at the University of Berlin.

Although his health was poor from about 1850 onward, he continued to lecture until 1890. His mathematical interests included algebraic functions, elliptic functions, geometry, and calculus.

## Setting Limits with Epsilon and Delta

“Delta” and “epsilon” are the fourth and fifth letters of the Greek alphabet. “Delta” often means “a change”, especially a small change in an independent variable. Here we use the lower-case “delta”, ‘δ’, and “epsilon”, ‘ε’, as the changes in ‘x’ and ‘y’ respectively when y=f(x).

We also need the “absolute value” function, represented by vertical bars. For example, “|a-b|” means “the absolute value of (a-b)”. This is always a positive real number, even if a<b.

Weierstrass proposed the delta-epsilon methodology for finding the limit of a function’s value as the independent variable approaches a value ‘X’. “Let y=f(x) be a real function. The limit of f(x) as x approaches ‘X’ exists and has the value ‘F’ if “|f(x) – F| < ε” whenever “zero < |x – X| < δ”.

In English, this would mean that the value of the function stays very close to the “limit” value as the independent variable gets very close to its specific value. Let’s work through an example.

## A Simple Example of How to Find a Mathematical Limit

The factor “(x – 1)/(x – 1)” has a constant value of ‘1’ except for the exact point where x=1, where it is undefined because the denominator, “x-1″ equals zero. Remember the first paragraph in this article?

The equation y = x + 1 is pretty dull, so let’s spice it up by multiplying by “(x – 1)/(x – 1)”. So we will deal with y = (x + 1)*(x – 1)/(x – 1).

Although we’ve noticed that ‘y’ is not defined at “x = 1″, we can determine the limit of ‘y’ as ‘x’ approaches the value ‘1’. The graph in the image shows that the straight line heading for (1, 2).

When ‘x’ has the value “(1 – δ)”, then y = ((1 – δ) + 1)*((1 – δ) – 1)/((1 – δ) – 1). That has a lot of ‘1’s to be simplified: y = ((2 – δ)*(δ))/(δ). As long as ‘δ’ is not zero, we can cancel out the δ/δ factor, leaving y = (2 – δ). Since ‘δ’ is a small real number, that means that the limit will be y=2.

We can make ‘δ’ as small as desired, but as long as it is not exactly zero, the function is defined and ‘y’ will have a value as close to ‘2’ as desired.

## Are there Limits to Limits?

Finding the limit of a function is not limited to regions where the function is undefined. In the above example, the limit of ‘y’ as “x approaches 2″ is easy to calculate: y = ((2 – δ) + 1)*((2 – δ) – 1)/((2 – δ) – 1) = (3 – δ)*(1 – δ)/(1 – δ) = (3 – δ) = 3 as ‘δ’ approaches zero. We could, of course, have simply plugged ‘2’ into the equation directly.

We’ve discussed an equation where the undefined region is “zero divided by zero”. Equations that are undefined for “infinity divided by infinity” might also be solved with a “limits” approach.

There are indeed limits to limits, however. Sometimes one is restricted to finding a “lower” or “upper” limit. One example is “y = (x – 1)/(square root of (x – 1)”. If we restrict the solution to real numbers rather than imaginary numbers, we could only allow ‘x’ to approach ‘1’ from above.

If the function oscillates between vastly different values, then there would be no one limit. An example is “y = 1/x”. As ‘x’ approaches zero from above, ‘y’ becomes a very large number. As ‘x’ approaches zero from below, ‘y’ becomes a very large negative number.

## Should Zeno Have Set Limits?

Two weeks ago, we resolved “Zeno’s Paradox of Achilles and the Tortoise” by using algebra. Zeno had set up the problem that Achilles can run ten times faster than the tortoise, but the tortoise has a 100-meter head start. Achilles repeatedly runs to the position where the tortoise **had been**. In that time the tortoise goes one-tenth of the distance farther. Therefore Achilles can never catch the tortoise.

Zeno wanted other Greek philosophers and mathematicians to fail, so he posed the question in terms that we could now describe as limits. The solution would be “How far does Achilles run so the remaining gap between Achilles and the tortoise approaches zero”?

Since this article has dragged on at some length, please trust that we could define Zeno’s problem as “a=f(g)”. Let ‘a’ represent the distance Achilles must have run in order to bring the gap, ‘g’, to zero. Remember that the gap is 100 meters before Achilles starts to run; bringing the gap to zero means Achilles must have run 100 meters.

Without showing the derivation of this new equation from the one found in the previous article, we need to solve “a = (1000 – g*10)/9″.

Let’s confirm this equation is accurate. At the start, when the gap, ‘g’, is 100 meters, a = (1000 – (100)*10)/9 = (1000 – 1000)/9 = zero. Yes, Achilles has run zero meters at the start. When the gap is 10 meters, a = (1000 – (10)*10)/9 = (1000 – 100)/9 = 900/9 = 100. Yes, Achilles did run 100 meters to reduce the gap to 10 meters.

We could simply plug “g = zero” into this equation: a = (1000 – (zero)*10)/9 = 1000/9 = 111.111… However, let’s try using the limit as the gap, ‘g’, approaches zero.

Then we want to find the limit of “a = (1000 – g*10)/9″ as ‘g’ approaches zero. We need to bring back our new friend ‘δ’. Then the equation is “a = (1000 – (zero + d)*10)/9″ = (1000 – d*10)/9. Obviously there is no problem in allowing ‘δ’ to become very close to zero. Therefore the solution, using a “limits” approach, is also 1000/9 = 111.111…

Zeno was probably just as happy that Weierstrass had not yet formalized the epsilon-delta method, since Zeno was trying to prove that motion was a paradox. He may not have welcomed any of the solutions modern mathematics could have provided.

**References**:

Friedman, E. Karl Theodor Wilhelm Weierstrass. Stetson. Referenced July 25, 2011.

Schechter, E. What are the “real numbers,” really?. Vanderbilt Mathematics. Nov. 11, 2009. Referenced July 25, 2011.

Weisstein, E. Limit. MathWorld, a Wolfram Web Resource. Referenced July 25, 2011.

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