Evaluating Limits Algebraically

Start with direct substitution

If f is a polynomial, rational (with nonzero denominator), root, exponential or trigonometric function continuous at a, then

lim (x→a) f(x) = f(a)

Try substitution first. If you get a real number, you are done.

When you get 0/0

The form 0/0 is indeterminate: it tells you nothing yet. Rewrite the expression.

1. Factor and cancel.

lim (x→3) (x² − 9)/(x − 3) = lim (x→3) (x − 3)(x + 3)/(x − 3) = lim (x→3) (x + 3) = 6

2. Rationalize (multiply by the conjugate).

lim (x→0) (√(x + 4) − 2)/x. Multiply top and bottom by √(x + 4) + 2:

= lim (x→0) x / (x(√(x + 4) + 2)) = 1/(2 + 2) = 1/4

3. Simplify complex fractions. Combine the small fractions over a common denominator, then cancel.

When you get a nonzero number over 0

For example 5/0. The limit is not a finite number: check one-sided behavior. It may be ∞, −∞, or DNE if the sides disagree.

Special trigonometric limits

• lim (x→0) sin x / x = 1
• lim (x→0) (1 − cos x)/x = 0

Adjust the argument: lim (x→0) sin(5x)/x = 5 · lim (x→0) sin(5x)/(5x) = 5.

Limit laws

Limits of sums, differences, products, quotients (nonzero denominator) and constant multiples can be split into separate limits.

The squeeze theorem

If g(x) ≤ f(x) ≤ h(x) near a, and lim g(x) = lim h(x) = L as x→a, then lim f(x) = L.

Classic use: since −x² ≤ x² sin(1/x) ≤ x², and both bounds go to 0, lim (x→0) x² sin(1/x) = 0.

Quick summary

• Substitute first.
• 0/0 means rewrite: factor, rationalize or simplify.
• Nonzero over zero means check for infinite limits.
• Memorize sin x / x → 1 and (1 − cos x)/x → 0.