The Limit Laws

The limit of a sum

Suppose f(x) approaches L and g(x) approaches M as $x\to c$x → c. Then f(x) + g(x) approaches L + M. The limit of a sum is the sum of the limits, so each piece may be taken separately.

Take $x^2+3x$x² + 3x as $x\to 2$x → 2. The piece $x^2$x² approaches 4 and the piece 3x approaches 6, so the sum approaches 4 + 6 = 10. Check near 2: at x = 1.99, $x^2=3.9601$x² = 3.9601 and 3x = 5.97, and the sum is 9.9301; at x = 2.01, $x^2=4.0401$x² = 4.0401 and 3x = 6.03, and the sum is 10.0701.

The same holds for a difference, f − g approaches L − M, and for a constant multiple, k f approaches kL.

$y=x^2$y = x² and y = 3x, and their sum $y=x^2+3x$y = x² + 3x in gold. Near x = 2 the two pieces head to 4 and 6, and the sum heads to 4 + 6 = 10.

Products, powers and polynomials

If f(x) approaches L and g(x) approaches M, then f(x) × g(x) approaches L × M. For x(x − 1) as $x\to 3$x → 3, x approaches 3 and x − 1 approaches 2, so the product approaches 6. Check: at 2.99 it is 5.9501, and at 3.01 it is 6.0501.

Two limits are known without any law: a constant k approaches k, and x approaches c as $x\to c$x → c. A power is a repeated product, so $x^2$x² approaches $c^2$c², $x^3$x³ approaches $c^3$c³, and so on. With the sum and constant-multiple laws, every polynomial follows: its limit as $x\to c$x → c is its value at c.

For example, $x^3-2x+5$x³ − 2x + 5 as $x\to 2$x → 2 approaches 8 − 4 + 5 = 9. Check: at 1.99 it is about 8.9006, and at 2.01 about 9.1006.

Quotients, with one condition

If f(x) approaches L and g(x) approaches M, and $M\ne 0$M ≠ 0, then $\frac{f\left(x\right)}{g\left(x\right)}$f(x)/g(x) approaches $\frac{L}{M}$L/M.

For $\frac{x^2+1}{x+3}$(x² + 1)/(x + 3) as $x\to 1$x → 1, the top approaches 2 and the bottom approaches 4, which is not 0, so the quotient approaches $\frac{2}{4}=\frac{1}{2}$2/4 = 1/2. Check: at 0.99 it is about 0.4963, and at 1.01 about 0.5038.

When the denominator tends to 0

If M = 0 the law cannot be used: it would divide by 0. Two different things can then happen.

If the top tends to a number other than 0, the quotient has no limit. $\frac{1}{x-2}$1/(x − 2) as $x\to 2$x → 2: the top is 1 and the bottom tends to 0. At 2.01 the quotient is 100 and at 2.001 it is 1000; at 1.99 it is −100 and at 1.999 it is −1000. It grows without bound on each side.

If the top tends to 0 as well, as in $\frac{x^2-4}{x-2}$(x² − 4)/(x − 2) as $x\to 2$x → 2, the law gives $\frac{0}{0}$0/0 and says nothing. The limit may still exist: this one is 4, the hole in The Idea of a Limit. Indeterminate Forms rewrites such a quotient so that the laws can be used again.

$y=\frac{1}{x-2}$y = 1/(x − 2). The top stays at 1 while the bottom tends to 0, so near x = 2 the curve runs off the top of the board on the right and off the bottom on the left. There is no limit at 2.

The laws need the limits to exist

Each law starts from limits of f and g that exist. Without them the law says nothing. Take f(x) = −1 for x < 0 and 1 for $x\ge 0$x ≥ 0, the step from One Sided Limits, and g(x) = −f(x). Neither has a limit at 0, but f(x) + g(x) = 0 for every x, which has limit 0.

So a sum can have a limit when its pieces do not. The laws go one way: from the limits of the pieces to the limit of the whole.

The usual mistakes

Mixing up the laws. If $f\to 3$f → 3 and $g\to 5$g → 5, then $f+g\to 8$f + g → 8 and $f\times g\to 15$f × g → 15: a sum adds the limits and a product multiplies them.

Using one limit alone. f × g needs the limit of g too; $f\to 3$f → 3 gives only one factor of the answer.

Using the quotient law when the denominator tends to 0. With $f\to 4$f → 4 and $g\to 0$g → 0, the law would divide 4 by 0, which is not a number; $f/g$f / g grows without bound.

Reading $\frac{0}{0}$0/0 as 0, or as no limit. It means the law gave no answer, and the quotient must be rewritten first.

A courier’s two-part charge

In the application below, each piece of the charge is a polynomial in w, so each one-sided limit at 10 kg is found by substituting w = 10 into its piece. Setting the two equal removes the jump.