Limits of algebraic functions

For the functions in this lesson there is a simple order of attack.

1. Try putting the number in. If it gives an ordinary answer, that is the limit.

• Polynomials: lim (x→2) (x³ – 2x + 1) = 8 – 4 + 1 = 5.
• Rational functions with a non-zero bottom: lim (x→1) (x² + 3)/(x + 1) = 4/2 = 2.
• A zero on top only is fine: the answer is simply 0.

2. If you get 0/0, change the form first. 0/0 is called an indeterminate form: it has no single value (0 = 0 × k for every k), so it tells you nothing yet. Something is hiding a common factor; remove it.

Factorise and cancel.

• lim (x→3) (x² – 5x + 6)/(x² – 9) = lim (x – 2)(x – 3)/[(x – 3)(x + 3)] = lim (x – 2)/(x + 3) = 1/6.
• The cancelling is allowed because x is near 3 but not equal to 3.

Combine fractions. lim (x→2) [1/(x – 2) – 4/(x² – 4)]: over x² – 4 this is (x + 2 – 4)/(x² – 4) = (x – 2)/[(x – 2)(x + 2)] = 1/(x + 2), which tends to 1/4.

Rationalise when there is a square root: lim (x→0) (√(4 + x) – 2)/x. Multiply top and bottom by √(4 + x) + 2: the top becomes (4 + x) – 4 = x, so the expression is 1/(√(4 + x) + 2), which tends to 1/4.

3. Use the standard result:

• lim (x→a) (xⁿ – aⁿ)/(x – a) = naⁿ⁻¹. It also works for fractional n when a > 0.
• lim (x→2) (x⁶ – 64)/(x – 2): 64 = 2⁶, so the answer is 6 × 2⁵ = 192.
• Two of them in a fraction: lim (x→2) (x⁶ – 64)/(x³ – 8). Divide top and bottom by (x – 2): (6 × 2⁵)/(3 × 2²) = 192/12 = 16.

A substitution can reveal the pattern. lim (x→0) ((1 + x)⁴ – 1)/x: put y = 1 + x, so y → 1 and the limit is (y⁴ – 1⁴)/(y – 1), which is 4 × 1³ = 4.

Only the bottom is 0? Then there is no limit. lim (x→1) (x + 1)/(x – 1): the top tends to 2 and the bottom to 0, so the fraction grows without bound (very negative from the left, very positive from the right).

Pieces. For a function defined in pieces, find the left limit with the left formula and the right limit with the right formula, and compare.