Closing In From Both Sides Limits and Derivatives

Suppose you drop a stone down a deep, dry well and watch it fall. Physics tells us the distance it has fallen after t seconds is s = 5t squared metres (taking the acceleration due to gravity as a tidy 10 m/s squared, half of which is 5). You want to know exactly how fast the stone is falling at the instant t = 3 seconds. Not its average speed over some stretch of the fall. Its speed right then, at that one instant. Here is the trouble. Speed is distance divided by time, and 'time' in that formula has to be an interval, a chunk with a beginning and an end. At a single instant, no time passes at all, so there is no interval to divide by. Ask for the speed 'at t = 3 seconds' literally, and you seem to be asking for a distance of zero divided by a time of zero, which is nonsense. And yet every driver's speedometer claims to show exactly this: speed at this instant. So either speedometers are lying, or there is a way to make sense of an instant's speed after all.

Here is the trick. Instead of asking for the speed at exactly t = 3, compute the average speed over short intervals that end at t = 3, and separately over short intervals that start at t = 3, and watch what happens as those intervals shrink. Average speed from t = 2.9 to t = 3: the stone falls s(3) minus s(2.9) = 45 minus 42.05 = 2.95 metres in 0.1 seconds, an average of 29.5 m/s. Shrink the interval to t = 2.99 to t = 3: the average speed works out to 29.95 m/s. Shrink it further, to t = 2.999 to t = 3, and the average climbs to 29.995 m/s. The numbers are marching steadily toward 30. Now approach from the other side. Average speed from t = 3 to t = 3.1: the stone falls s(3.1) minus s(3) = 48.05 minus 45 = 3.05 metres in 0.1 seconds, an average of 30.5 m/s. Shrink to t = 3 to t = 3.01: average speed 30.05 m/s. Shrink to t = 3 to t = 3.001: average speed 30.005 m/s. This second sequence is marching down toward 30 as well. Both sequences, approaching t = 3 from before and from after, are closing in on the exact same number, 30 m/s, even though neither sequence ever actually reaches an interval of zero length. That shared target is what we mean by the instantaneous speed at t = 3. Not a measurement. A destination that both directions of approach agree on.

What we just did to the falling stone, we can do to any function. Given a function f(x) and a target input a, we ask what value f(x) approaches as x is pushed closer and closer to a, without ever actually setting x equal to a. If the values of f(x) approach some single number L as x approaches a from values less than a (called the left hand limit, written lim as x to a from the left of f(x)), and f(x) also approaches that same L as x approaches a from values greater than a (the right hand limit), then L is called the limit of f(x) as x tends to a, written lim(x to a) f(x) = L. The two one-sided limits have to agree for the limit to exist at all. Try f(x) = x + 5 at a = 2. Values of x just below 2, like 1.9, 1.99, 1.999, give f(x) values of 6.9, 6.99, 6.999, climbing toward 7. Values of x just above 2, like 2.1, 2.01, 2.001, give f(x) values of 7.1, 7.01, 7.001, also closing in on 7 from above. Both sides agree: lim(x to 2)(x + 5) = 7. Here the limit even matches the plain function value f(2) = 7, but that match is not guaranteed and should never be assumed. Consider g(x) = (x squared minus 9)/(x minus 3). At x = 3 exactly, g is not even defined, since we would be dividing by zero. But for any x close to 3 and not equal to it, (x squared minus 9)/(x minus 3) simplifies to (x + 3), because x squared minus 9 factors as (x minus 3)(x + 3) and the (x minus 3) cancels. So as x approaches 3 from either side, g(x) behaves exactly like x + 3, and the limit is 3 + 3 = 6, even though g(3) itself does not exist. A limit describes the trend a function is heading toward, not necessarily the value it actually lands on.

Sometimes the left hand limit and the right hand limit are both perfectly well defined, and simply different, in which case the two-sided limit does not exist at all. A parking garage might charge a flat 40 rupees for up to one hour, then jump to 70 rupees for anything over one hour, no matter how small the excess. Model the cost as a function of parked-time t, in hours: c(t) = 40 for t less than or equal to 1, and c(t) = 70 for t greater than 1. Approaching t = 1 from below, at t = 0.9, 0.99, 0.999 hours, the cost is 40 rupees throughout, so the left hand limit is 40. Approaching t = 1 from above, at t = 1.1, 1.01, 1.001 hours, the cost is 70 rupees throughout, so the right hand limit is 70. The two one-sided limits are both perfectly definite, and they disagree, 40 against 70, so lim(t to 1) c(t) simply does not exist. This is not a failure of the method. It is the method correctly reporting that the function genuinely jumps at t = 1, and no single number can honestly describe what c(t) is 'heading toward' there, because it is heading toward two different places depending on which direction you approach from.

One pair of limits turns out to matter more than any other in this entire subject: lim(x to 0) of sin(x)/x, and lim(x to 0) of (1 - cos x)/x. Both x = 0 substitutions give the meaningless 0/0, so direct substitution fails, and factoring will not help either since sin x is not a polynomial. The way through is a squeeze. Picture a circle of radius 1, and a small angle x (in radians) at its centre. Three areas sit inside one another for small positive x: the triangle formed by the centre and the chord, the pie-slice (sector) of the circle itself, and a larger triangle formed using the tangent line. Comparing these three areas geometrically gives the inequality cos(x) is less than sin(x)/x, which in turn is less than 1, for small x. As x is squeezed toward 0, cos(x) itself approaches 1 (cosine is a smooth, continuous function equal to 1 at x=0), and the constant 1 on the other side is obviously always 1. Sin(x)/x is trapped between two things that are both closing in on 1, so it has no choice but to close in on 1 as well. This squeezing argument, formally called the Sandwich Theorem, is what gives lim(x to 0) sin(x)/x = 1. A companion identity, 1 - cos(x) = 2 sin squared(x/2), turns the second limit into a rearrangement of the first, giving lim(x to 0) (1 - cos x)/x = 0. These two results look narrow, but they are the engine behind every derivative of sin, cos, and tan you will ever compute.

Return to the falling stone. Instead of the position function s(t) = 5t squared, write a general function f(x), and instead of one specific instant, ask for the instantaneous rate of change at any point x. The average rate of change of f between x and a nearby point x + h is [f(x+h) minus f(x)] divided by h, exactly the average-speed calculation from before, generalised. As h shrinks toward 0, this average rate of change approaches a limiting value, provided that limit exists. That limiting value is called the derivative of f at x, written f prime of x, or df/dx: f prime(x) = lim(h to 0) [f(x+h) minus f(x)]/h. Geometrically, [f(x+h) minus f(x)]/h is the slope of the straight line joining the two points (x, f(x)) and (x+h, f(x+h)) on the graph of f, called a secant line. As h shrinks to 0, the second point slides along the curve toward the first, and the secant line rotates into the tangent line, the single straight line that just grazes the curve at that one point without crossing through it locally. The derivative f prime(x) is exactly the slope of that tangent. Apply the definition to f(x) = x squared. f prime(x) = lim(h to 0) [(x+h) squared minus x squared]/h = lim(h to 0) [2xh + h squared]/h = lim(h to 0) (2x + h) = 2x, since as h shrinks to 0 the leftover h term simply vanishes. So the derivative of x squared is 2x, meaning the slope of the parabola y = x squared at any point x is exactly twice that x-coordinate. At x = 3, the slope is 6; at x = negative 1, the slope is negative 2.

Computing every derivative from the raw limit definition, as we just did for x squared, works, but it is slow. A short list of rules, each itself provable from the definition, lets you differentiate almost anything quickly. The power rule: the derivative of x to the power n is n times x to the power (n minus 1), for any real number n. The sum and difference rules: the derivative of f(x) plus or minus g(x) is f prime(x) plus or minus g prime(x); derivatives distribute over addition and subtraction exactly like limits do. The product rule (sometimes called the Leibnitz rule): the derivative of f(x) times g(x) is f prime(x) times g(x), plus f(x) times g prime(x). Note this is not simply f prime times g prime; the two terms both matter. The quotient rule: the derivative of f(x)/g(x) is [f prime(x) times g(x), minus f(x) times g prime(x)], all divided by g(x) squared, valid wherever g(x) is not zero. Finally, three standard trigonometric derivatives, each derivable from the two special limits above: the derivative of sin(x) is cos(x); the derivative of cos(x) is negative sin(x); the derivative of tan(x) is sec squared(x). Notice the pattern with sine and cosine: differentiating sin gives cos, and differentiating cos gives negative sin, a pair that will keep reappearing throughout this whole thread of chapters whenever oscillating or repeating behaviour needs describing.

Nothing in the whole Calculus thread before Class 11 exists. This is the true starting point. Every idea in the five chapters that follow, continuity, rates of change, maxima and minima, integrals, and differential equations, is built directly on the two tools defined here: the limit, and the derivative as a limit of slopes. When a later chapter says a function is continuous, it is making a precise claim using limits. When a later chapter finds the fastest a tank can be filled, or the steepest point on a curve, it is using the derivative rules from this chapter, just applied to new questions. Two mathematicians are usually credited with inventing calculus independently around the same time in the seventeenth century, Isaac Newton in England and Gottfried Leibniz in Germany, each developing their own notation and each motivated by a different problem (Newton by motion and physics, Leibniz more by the geometry of curves). Newton's fluxions notation has mostly disappeared, but Leibniz's d/dx notation, used throughout this chapter, is still the one written on every exam paper today. It took another century and a half, and the careful work of mathematicians including Augustin-Louis Cauchy, to turn the intuitive 'gets closer and closer' language used throughout this chapter into the fully rigorous definition of a limit used in university mathematics, built from precise inequalities rather than tables of approaching numbers. What you have built here, informal and table-driven as it is, is exactly the right level to start from.