The Idea of a Limit

Closing in on 2

Start at 1 and go halfway to 2: 1.5. Go halfway again: 1.75. Then 1.875, then 1.9375. The gap to 2 is 1, then 0.5, 0.25, 0.125, 0.0625, halving every time.

No step lands on 2, because half of a gap is never zero. But the values get as close to 2 as you like: after ten steps the gap is $\frac{1}{1024}$1/1024, less than 0.001. That is what it means to say the values approach 2, and 2 is their limit.

The values 1, 1.5, 1.75, 1.875 and 1.9375, each halfway from the one before to 2. They crowd toward the gold mark at 2 and never reach it.

Near 2, not at 2

For a function, the question is what f(x) approaches as x approaches a number c. Saying that the limit of f(x) as $x\to c$x → c is L means that f(x) can be made as close to L as you like by taking x close enough to c, from either side, with $x\ne c$x ≠ c.

Take f(x) = x + 1 and c = 2. At x = 1.9, 1.99 and 1.999, f(x) is 2.9, 2.99 and 2.999. At x = 2.1, 2.01 and 2.001 it is 3.1, 3.01 and 3.001. From both sides the outputs close in on 3, so the limit of x + 1 as $x\to 2$x → 2 is 3.

Here f(2) = 3 as well, but the limit did not use it. Only inputs near 2 were tried, never 2 itself.

f(x) = x + 1 at inputs below 2 and above 2. Below, the outputs climb toward 3; above, they fall toward 3. No column is x = 2.

A function with a hole

Now take $f\left(x\right)=\frac{x^2-4}{x-2}$f(x) = (x² − 4)/(x − 2). At x = 2 the top and the bottom are both 0, and $\frac{0}{0}$0/0 is not a number, so f(2) is undefined.

Near 2 it is perfectly well behaved: f(1.9) = 3.9, f(1.99) = 3.99, f(1.999) = 3.999, and f(2.1) = 4.1, f(2.01) = 4.01, f(2.001) = 4.001. The outputs approach 4 from both sides, so the limit of $\frac{x^2-4}{x-2}$(x² − 4)/(x − 2) as $x\to 2$x → 2 is 4.

The reason: $x^2-4=\left(x+2\right)\left(x-2\right)$x² − 4 = (x + 2)(x − 2), so for every x except 2 the factor x − 2 cancels and f(x) = x + 2. The graph is the line y = x + 2 with one point missing, at (2, 4). The limit asks only about points near 2, so the missing point does not stop it.

The values of $\frac{x^2-4}{x-2}$(x² − 4)/(x − 2) close in on 4 from below and from above. At x = 2 itself there is no value.

The graph of $\frac{x^2-4}{x-2}$(x² − 4)/(x − 2): the line y = x + 2 with a hole at (2, 4). The line runs into the hole from both sides.

The value at the point plays no part

Define g(x) = x + 2 for $x\ne 2$x ≠ 2, and g(2) = 7. Near 2, g agrees with x + 2, so its outputs still approach 4: the limit of g(x) as $x\to 2$x → 2 is 4, even though g(2) = 7.

So a limit can exist where the function has no value, and where it has a different value. The limit says where the outputs are heading; the value says where the function actually is at that one input.

g(x) = x + 2 everywhere except x = 2, where g(2) = 7. The line still runs into the hole at (2, 4), so the limit at 2 is 4; the filled point at (2, 7) is the value.

When there is no limit

A limit needs one number for the outputs to approach. $h\left(x\right)=\frac{1}{x^2}$h(x) = 1/x² has none at x = 0: h(0.1) = 100, h(0.01) = 10 000 and h(0.001) = 1 000 000, and the negative inputs −0.1 and −0.01 give the same values. The outputs grow without bound, past every number you name, so $\frac{1}{x^2}$1/x² has no limit as $x\to 0$x → 0.

A function can also fail by heading to two different numbers from the two sides. That is the subject of One Sided Limits.

Limits of sequences

The same idea works for a list of numbers as n grows without bound. $\frac{1}{n}$1/n is 0.1 at n = 10, 0.01 at n = 100 and 0.001 at n = 1000: it shrinks toward 0, so its limit is 0.

$2+\frac{1}{n}$2 + 1/n approaches 2, staying just above it. $\frac{3n+1}{n}$(3n + 1)/n splits as $3+\frac{1}{n}$3 + 1/n, which is 3.1, 3.01, 3.001 at n = 10, 100, 1000, and approaches 3.

The running totals of $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots$1/2 + 1/4 + 1/8 + … are $\frac{1}{2}$1/2, $\frac{3}{4}$3/4, $\frac{7}{8}$7/8, $\frac{15}{16}$15/16, …: each total is 1 minus the latest term, so the gap to 1 halves every time. The sum approaches 1, and continuing forever, it is 1.

The usual mistakes

Saying there is no limit because the function is undefined at the point. $\frac{x^2-4}{x-2}$(x² − 4)/(x − 2) has no value at 2, and its limit there is 4.

Taking the value at the point as the limit. g(2) = 7, but the outputs near 2 approach 4.

Reading the numerator as the limit of $\frac{1}{n}$1/n. The fraction keeps shrinking, $\frac{1}{10}$1/10, $\frac{1}{100}$1/100, $\frac{1}{1000}$1/1000, so its limit is 0, not 1.

Adding the pieces of $\frac{3n+1}{n}$(3n + 1)/n. The 1 is divided by n, so it shrinks away and the limit is 3, not 4.

Giving 2 for $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots$1/2 + 1/4 + 1/8 + … . The totals climb toward 1 and never pass it; a sum reaching 2 would have to start with a term of 1.