Factoring, Rationalizing and Squeezing Limits

Diagnose before calculating

Direct substitution is the first check when a formula is continuous at the target. If it produces $0/0$ , both numerator and denominator are tending to zero, and their relative rates matter. Factoring and rationalizing reveal a nearby equivalent expression. Squeezing controls a difficult expression between two simpler ones.

An algebraic replacement need only agree in a punctured neighborhood of the target. It may have a different value or domain exactly at the target, because a limit ignores that single point.

Visual guide

VISUAL GUIDECancellation leaves a hole

Quotient for x ≠ 1

For x ≠ 1, (x² − 1)/(x − 1) equals x + 1. The open circle marks the missing value at x = 1; the nearby graph still approaches 2 from both sides.

Worked example: conjugates

To find $\lim_{x\to0}(\sqrt{1+x}-1)/x$ , multiply by the conjugate:

\frac{\sqrt{1+x}-1}{x}=\frac{1}{\sqrt{1+x}+1},

x\ne0.

The right-hand expression is continuous near zero and approaches $1/2$ . Multiplying only the numerator would change the expression; use the same nonzero factor above and below. The square-root domain also requires $x\ge-1$ .

Worked example: control every approach

Since $-1\le\sin(1/x)\le1$ for nonzero $x$ , multiplication by $x^2\ge0$ gives

-x^2\le x^2\sin(1/x)\le x^2.

Both outside expressions approach zero, so the middle expression does too. The oscillating factor alone has no limit, but its amplitude becomes small enough to force a limit for the product.

The squeeze theorem requires the inequalities to hold throughout a sufficiently small neighborhood. Checking a few inputs, or comparing two selected paths, does not establish that condition.

A fundamental trigonometric limit

With angles in radians, $\lim_{u\to0}\sin u/u=1$ . A geometric squeeze on the unit circle establishes this before derivatives are known. Thus

\lim_{x\to0}\frac{\sin(5x)}{2x}=\frac52\lim_{x\to0}\frac{\sin(5x)}{5x}=\frac52.

The matching inner argument matters. Also, using $1-\cos x=2\sin^2(x/2)$ gives $(1-\cos x)/x^2\to1/2$ . Degree-based trigonometric functions introduce a conversion factor, so the radian assumption is essential.

Choosing the next step

Polynomial differences often factor. Differences involving roots often invite a conjugate. A bounded oscillation multiplied by something tending to zero often invites squeezing. Do not use L’Hôpital's rule before establishing its hypotheses, and avoid using it to prove a trigonometric limit on which your derivative formulas already depend.

Practice

Evaluate $\lim_{x\to4}(\sqrt{x}-2)/(x-4)$ .
Evaluate $\lim_{x\to0}x\cos(1/x)$ .
Evaluate $\lim_{x\to0}\sin(3x)/\sin(2x)$ .

Show worked solutions

Rationalization gives $1/(\sqrt{x}+2)$ , hence $1/4$ .
Its absolute value is at most $|x|$ , so the limit is zero. This bound handles both signs of $x$ .
Write the ratio as $[\sin(3x)/(3x)]/[\sin(2x)/(2x)]\cdot3/2$ . Both bracketed limits are one, giving $3/2$ .