Topic 1.6 Determining Limits Using Algebraic Manipulation: resolve indeterminate forms by factoring and cancelling, rationalizing, combining fractions, or using known trigonometric limits.

What this topic is asking

The College Board (Topic 1.6) wants you to evaluate limits where direct substitution gives the indeterminate form $\frac{0}{0}$. The fix is to rewrite the function into an equivalent form (valid for $x \neq a$) on which substitution does work. The standard tools are factoring and cancelling, rationalizing with a conjugate, combining fractions, and a few known trigonometric limits.

Why $\frac{0}{0}$ means "keep going"

Technique 1: factor and cancel

For rational expressions, factor numerator and denominator and cancel the factor that is zero at $a$.

Technique 2: rationalize with a conjugate

When a square root creates the $\frac{0}{0}$, multiply top and bottom by the conjugate to turn the root into a difference of squares.

Technique 3: combine a complex fraction

When the expression has fractions stacked inside fractions, combine them over a common denominator first.

Known trigonometric limits

You often rewrite a trigonometric limit until one of these appears, for example $\lim_{x \to 0} \frac{\sin 5x}{x} = \lim_{x \to 0} 5 \cdot \frac{\sin 5x}{5x} = 5(1) = 5$.

Why rewriting is legal

A reasonable worry is whether you are "allowed" to cancel a factor that is zero at the very point you care about. The answer is yes, precisely because a limit ignores the single point $x = a$. When you cancel $(x - a)$ from $\frac{(x-a)(x+2)}{(x-a)(x-2)}$, the simplified function $\frac{x+2}{x-2}$ disagrees with the original only at $x = a$, where the original was undefined anyway. Everywhere else - including every point in the approach toward $a$ - the two functions are identical, so they must have the same limit. This is the formal justification for every algebraic rewrite in the topic: the new expression is a different function (it may be defined where the old one was not), but it shares all the nearby values that determine the limit. Stating this once to yourself dissolves the discomfort and explains why substitution into the simplified form gives the right answer.

Choosing the right tool

Match the obstacle to the technique: a factorable polynomial calls for factor-and-cancel; a root calls for the conjugate; stacked fractions call for combining; a trigonometric ratio calls for the special limits. If substitution gives a nonzero number over zero instead of $\frac{0}{0}$, the limit is infinite and these techniques do not apply. Occasionally a single limit needs two tools in sequence - for instance combining a complex fraction and then cancelling, or rationalizing and then factoring - so do not assume one move always finishes the job. After each rewrite, substitute again to check whether you have escaped the indeterminate form; if you are still at $\frac{0}{0}$, keep simplifying. This substitute-rewrite-resubstitute loop is the dependable rhythm for every algebraic limit on the exam.