Limits

Closing in on 2

Start at 1, where the gap to 2 is 1. Close half of the gap: 1.5. Close half of what is left: 1.75. Then 1.875, then 1.9375. After n halvings the gap is $\frac{1}{2^n}$1/2ⁿ, so the numbers are $2-\frac{1}{2}$2 − 1/2, $2-\frac{1}{4}$2 − 1/4, $2-\frac{1}{8}$2 − 1/8, and so on.

No number in this list is 2, because there is always a gap left. But the gap can be made as small as you like: after 10 halvings it is $\frac{1}{1024}$1/1024, less than 0.001, and every later number is closer still. The numbers close in on 2, and 2 is called their limit.

From 1, each hop closes half of the gap that is left to 2: 1.5, 1.75, 1.875, 1.9375. The hops get shorter and never reach the gold mark at 2.

What a limit is

A sequence has the limit L when its terms get as close to L as you like, and stay that close, as you go further along. The terms do not have to reach L. The limit is written with an arrow: $2-\frac{1}{2^n}\to 2$2 − 1/2ⁿ → 2 as $n\to \infty$n → ∞, read "tends to 2 as n tends to infinity".

As n grows, $\frac{1}{n}$1/n tends to 0: its terms are 1, 0.1 at n = 10, 0.01 at n = 100, and they fall below any small number you name. Then $1+\frac{1}{n}$1 + 1/n tends to 1, because the 1 stays and the $\frac{1}{n}$1/n part tends to 0. And $\frac{\left(2n+1\right)}{n}$(2n + 1)/n is $2+\frac{1}{n}$2 + 1/n after dividing through by n, so it tends to 2.

An endless sum has a limit too. The running totals of $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots$1/2 + 1/4 + 1/8 + … are 0.5, 0.75, 0.875, 0.9375, …, and each one leaves a gap below 1 that is half the gap before. So the sum closes in on 1.

Down each column the terms close in on a limit: $\frac{1}{n}$1/n on 0, $1+\frac{1}{n}$1 + 1/n on 1, and $\frac{\left(2n+1\right)}{n}$(2n + 1)/n on 2.

A value the function never takes

A function can have a limit at a point where it has no value. Take $\frac{\left(x^2-4\right)}{\left(x-2\right)}$(x² − 4)/(x − 2). At x = 2 it is $\frac{0}{0}$0/0, which has no value. But for every other x the top factors as (x − 2)(x + 2), and the (x − 2) cancels, leaving x + 2.

So near 2 the function is close to 4: at x = 1.9 it is 3.9, at x = 1.99 it is 3.99, and at x = 2.01 it is 4.01. As x tends to 2 from either side, $\frac{\left(x^2-4\right)}{\left(x-2\right)}$(x² − 4)/(x − 2) tends to 4, even though it is never equal to 4. This is the kind of limit calculus needs.

The graph of $\frac{\left(x^2-4\right)}{\left(x-2\right)}$(x² − 4)/(x − 2) is the line y = x + 2 with one point missing, drawn as an open circle at (2, 4). The graph closes in on that point from both sides.

Chords on a curve

On $y=x^2$y = x², take the point (2, 4) and a second point h further along, at $\left(2+h,{\left(2+h\right)}^2\right)$(2 + h, (2 + h)²). The chord between them has gradient $\left({\left(2+h\right)}^2-4\right)/h$((2 + h)² − 4) / h.

With h = 1 the second point is (3, 9), and the gradient is (9 − 4) ÷ 1 = 5. With h = 0.5 it is (6.25 − 4) ÷ 0.5 = 4.5. With h = 0.1 it is (4.41 − 4) ÷ 0.1 = 4.1, and with h = 0.01 it is (4.0401 − 4) ÷ 0.01 = 4.01. The nearer the two points, the closer the chord lies to the tangent at (2, 4).

The chord through (2, 4) and (3, 9) on $y=x^2$y = x² has gradient 5. The gradient of the curve at (2, 4) is a little less.

The chord settles onto the tangent

Expand the top: ${\left(2+h\right)}^2-4=4h+h^2$(2 + h)² − 4 = 4h + h², so for any h that is not 0 the chord gradient is $\left(4h+h^2\right)/h=4+h$(4h + h²) / h = 4 + h. That matches every value above: 5, 4.5, 4.1, 4.01.

With h = 0 the two points are the same point, and the formula gives 0 ÷ 0, which has no value. But as h tends to 0, 4 + h tends to 4. The chords settle onto the tangent, and the gradient of the tangent at (2, 4) is the limit, 4.

Drag the upper point down the curve toward (2, 4). As the gap h shrinks, the dashed chord turns toward the tangent, and at h = 0 only the tangent is left, with gradient 4. The note under the drawing writes f'(2) for the gradient of the curve at x = 2.

From both sides

The second point can also be to the left of (2, 4). With h = −0.1 it is (1.9, 3.61), and the chord gradient is (3.61 − 4) ÷ (−0.1) = 3.9, which is 4 + h again. Chords from the left have gradients just under 4, and chords from the right just over 4. Both close in on 4, so the limit is 4.

The usual mistakes

Taking the first term for the limit. $\frac{1}{n}$1/n starts at 1, but its terms keep shrinking, so it closes in on 0.

Taking the limit of one part for the whole. In $1+\frac{1}{n}$1 + 1/n, the $\frac{1}{n}$1/n part tends to 0, but the 1 stays, so the whole tends to 1.

Thinking a sum of endlessly many positive terms must grow without end. Each term of $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots$1/2 + 1/4 + 1/8 + … fills half of the gap below 1, so the total never passes 1.

Putting h = 0 into $\left({\left(2+h\right)}^2-4\right)/h$((2 + h)² − 4) / h before simplifying, which gives 0 ÷ 0. Expand and divide by h first; then let h tend to 0.

A car pulling away

In the application below, the distance a car has traveled is $s=2x^2$s = 2x² meters after x seconds. Its average speeds over shorter and shorter intervals from x = 3 are chord gradients, and their limit is its speed at that instant.