⚖️Limit Laws & the Formal Definition

At Limit's Edge you have already learned what a limit *is*: the height a function closes in on, whether or not it ever arrives. This craft is about what a limit *obeys*. A limit turns out to be a very well-behaved object — it survives every ordinary operation of arithmetic — and the list of ways it does that is called the limit laws.

Suppose that as x approaches a, f(x) approaches p and g(x) approaches q. Then, as x approaches a:

• ◆Sum and difference. f(x) + g(x) approaches p + q, and f(x) - g(x) approaches p - q.
• ◆Constant multiple. c*f(x) approaches c*p. A constant walks straight through a limit.
• ◆Product. f(x)g(x) approaches p*q.
• ◆Quotient. f(x)/g(x) approaches p/q, provided q is not 0.
• ◆Power. (f(x))ⁿ approaches pⁿ. Squaring is the case you meet most.
• ◆Root. sqrt(f(x)) approaches sqrt(p), as long as p is not negative.

Read those as one sentence: *the limit of a combination is that same combination of the limits.* You take the expression apart along its operations, replace each piece by its own limit, and put it back together. That is the whole method, and it is why substitution works for every polynomial: a polynomial is built from x and constants using nothing but sums, constant multiples and products.

Two of the laws have small print, and the small print is where the marks are.

The quotient law refuses to work when the bottom limit is 0. If f approaches 5 and g approaches 0, then f/g is a number settling on 5 divided by a number shrinking to nothing, and that quotient does not settle anywhere at all — it runs away. The law gives no value, and no amount of algebra rescues it, because there is no hidden common factor to cancel: the top is not heading for 0. Notice how different this is from the 0/0 case you met in Limits, where both parts shrink and a cancellation is waiting. Check the bottom limit before you use the quotient law, every time.

The root law needs a non-negative limit inside, for the ordinary reason that sqrt of a negative number is not a real number.

There is one more law, and it is the one that makes composite functions work.

Composition. If g(x) approaches b as x approaches a, and f is continuous at b, then f(g(x)) approaches f(b). Work from the inside out: find where the inside settles, then apply the outside to that value. The word "continuous" is doing all the work — it is exactly the condition that lets you swap "the limit of f" for "f of the limit".

Substitution gives 0/0. In Limits you factored and cancelled. That works for polynomials, but three other shapes turn up constantly and each has its own move.

A square root on top or bottom: multiply by the conjugate. The conjugate of sqrt(A) - B is sqrt(A) + B, and multiplying them gives A - B² — the radical vanishes, because (u - v)(u + v) = u² - v². So for

(sqrt(x + 7) - 3) / (x - 2) as x approaches 2

multiply top and bottom by (sqrt(x + 7) + 3). The top becomes (x + 7) - 9 = x - 2, which cancels the bottom exactly, and what is left is

1 / (sqrt(x + 7) + 3) -> 1 / (3 + 3) = 1/6

The single most common slip here is stopping at 1/3: after the cancellation the denominator is sqrt(9) + 3, which is 3 + 3, not 3. And if the radical is on the *bottom* instead, the same conjugate clears it and the answer comes out bigger than 1 rather than smaller — the whole fraction is the other way up.

A fraction inside a fraction: combine the top over one denominator. For

(1/x - 1/5) / (x - 5) as x approaches 5

put the top over the common denominator 5x: it becomes (5 - x)/(5x). Now 5 - x is the negative of x - 5, so dividing by (x - 5) cancels that factor and leaves a minus sign behind:

-1 / (5x) -> -1/25

Two things to guard. The minus sign is not optional — it comes from the order of the subtraction, and dropping it is the classic error. And the answer has 5 twice underneath, not once: 5x becomes 5 times 5 when x arrives.

The same shape appears with h instead of a number:

(1/(x + h) - 1/x) / h as h approaches 0

Combine the top over x(x + h) to get -h/(x(x + h)), cancel the h, and let h go to 0: the answer is -1/(x²), an expression in x rather than a single number. (You may recognise it as the derivative of 1/x, which is exactly what it is.)

An absolute value: split into two one-sided questions. |x - a| is two rules wearing one coat: it equals x - a when x is above a, and -(x - a) when x is below. So a limit containing one has to be done twice.

|x - 4| / (x - 4)

is +1 for every x above 4 and -1 for every x below it. The limit from the right is 1, the limit from the left is -1, and because those disagree the two-sided limit does not exist. Nothing here is anywhere near 0, so 0 is never the answer; and "does not exist" is a statement about the two sides disagreeing, not about the function having no value at the point.

Substituting 0 into sin(x)/x gives 0/0, and there is nothing to factor and no radical to rationalise. Two limits have to be established once, geometrically, and then everything else is built from them. With x in radians:

sin(t)/t -> 1 as t approaches 0 (1 - cos(t))/t -> 0 as t approaches 0

The reason for the first, in one line: for a small angle t in a unit circle, the arc of length t, the chord and the vertical sin(t) are squeezed between each other, and the ratio of any two of them closes on 1. That is why radians matter — in degrees the arc is not t, and the limit would be pi/180 instead. The reason for the second, also in one line: multiply top and bottom by (1 + cos t) and the top becomes sin(t)², which shrinks like t² while the bottom shrinks like t.

Everything else is bookkeeping. The angle inside the sine and the thing underneath must match. They rarely do, so you make them match and pay for it with a factor in front:

sin(5x)/x = (5)(sin(5x)/(5x)) -> (5)(1) = 5

so sin(kx)/x approaches k — not 1. Dropping the k is the error this whole family is built to catch. The same trick handles every variation:

• ◆sin(kx)/(mx) = (k/m)(sin(kx)/(kx)) -> k/m
• ◆x/sin(kx) is the reciprocal, so it approaches 1/k
• ◆tan(kx)/x: write tan = sin/cos. The cosine approaches cos(0) = 1, so this behaves exactly like sin(kx)/x and approaches k
• ◆sin(ax)/sin(bx): divide top and bottom by x. The top approaches a, the bottom approaches b, and the quotient law gives a/b
• ◆(1 - cos(x))/x²: multiply top and bottom by (1 + cos x). The top becomes sin(x)², so the expression is (sin(x)/x)² divided by (1 + cos x), which approaches (1)(1)/2 = 1/2

That last one is worth memorising alongside the other two, because the two cosine limits have different values — 0 over x, and 1/2 over x² — and mixing them up is a favourite trap.

Sometimes you cannot get at f at all. You have no formula, or the formula misbehaves so badly that no law applies. The squeeze theorem (also called the sandwich theorem) gets an answer anyway, from two functions that trap it.

If g(x) <= f(x) <= h(x) for every x near a, and g and h both approach the same value L as x approaches a, then f approaches L too.

The picture is the whole proof. Draw two curves that pinch together at x = a, meeting at height L. Now draw anything at all between them. It has nowhere to go: as x closes on a the corridor narrows to a single point, and whatever is inside is dragged to that height. Note what the theorem does not need — any rule for f, any continuity of f, any value for f at a. Two walls and a squeeze are enough.

A textbook use. Suppose 4x - 9 <= f(x) <= x² - 4x + 7 for every x near 4. Substituting x = 4 into the lower bound gives 7; substituting into the upper bound gives 16 - 16 + 7 = 7. The two walls meet at 7, so the limit of f as x approaches 4 is 7 — and we still have no idea what f is.

The use it was invented for. Consider f(x) = x²*sin(1/x) for x other than 0. As x approaches 0, the angle 1/x races off, so sin(1/x) swings between -1 and 1 faster and faster and has no limit whatever. But it never leaves that band, and x² is never negative, so

-x² <= x²*sin(1/x) <= x²

Both walls approach 0, so the limit is 0. The swinging is crushed by the shrinking factor. Two warnings, both common: the sine reaching 1 somewhere does not make the limit 1, because it is still multiplied by something heading for 0; and the fact that sin(1/x) alone has no limit says nothing about the product. If the same idea is shifted — say 3 - x² <= f(x) <= 3 + x² — then the walls close on 3, and 3 is the answer. The squeeze theorem gives whatever the walls agree on, which is not always 0.

Everything above rests on the word "approaches", and so far that word has been an intuition. The formal definition makes it exact, and it is best read as a bargain between two people.

You say: *I want f(x) to be within a distance of L. Here is how close: epsilon.* You may choose epsilon as small as you like — a thousandth, a millionth, anything above 0.

I reply: *Then keep x within delta of a, and I promise you it will be.* My delta is allowed to depend on your epsilon, and it usually does; the promise covers every x within delta of a except a itself, because a limit never looks at the point.

The limit of f(x) as x approaches a is L means: for every epsilon greater than 0 there is a delta greater than 0 such that every x with 0 < |x - a| < delta has |f(x) - L| < epsilon.

The order of those two is the entire content of the definition. Epsilon is demanded first and delta answers it. Swap them and you have said something much weaker and quite useless. And "for every epsilon" cannot be relaxed to "for some epsilon", or a function sitting a fixed distance away from L would pass.

Finding delta for a straight line. This is one calculation, and it goes backwards from what you want. Take f(x) = 3x + 1, a = 2, so L = 7, and suppose epsilon = 0.6.

Start with the thing to be guaranteed and unwind it:

|f(x) - 7| < 0.6 |3x + 1 - 7| < 0.6 |3x - 6| < 0.6 3|x - 2| < 0.6 |x - 2| < 0.2

Every line is reversible, so |x - 2| < 0.2 guarantees |f(x) - 7| < 0.6. The largest window is delta = 0.2 = 1/5, and in general, for f(x) = mx + c,

delta = epsilon / |m|

The slope is a magnifier: a steep line turns a small wobble in x into a big wobble in f, so a steep line needs a narrow window. Setting delta = epsilon works only when |m| = 1, and it is the standard mistake. Multiplying instead of dividing is the same mistake wearing its coat inside out. The absolute value matters too — a window is a distance, so it is the *size* of the slope that divides.

The question also comes the other way round. Given a window, what tolerance does it buy? Then you run the same chain forwards: |x - a| < delta gives |f(x) - L| = |m||x - a| < |m|*delta. Going from x to f multiplies by |m|; going from f to x divides by it.