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AP Calculus AB: a complete guide to evaluating limits algebraically on the exam

A deep-dive AP Calculus AB guide to evaluating limits algebraically. Covers the substitution-first routine, resolving 0/0 forms by factoring, conjugates and combining fractions, the special trigonometric limits, infinite limits and limits at infinity, and the no-calculator exam technique the College Board rewards.

Generated by Claude Opus 4.818 min read1.5-1.16

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Jump to a section
  1. What evaluating limits actually demands
  2. The substitution-first routine
  3. Resolving 0/0 by rewriting
  4. The special trigonometric limits
  5. Infinite limits and sign analysis
  6. Limits at infinity
  7. How limit evaluation is examined
  8. Check your knowledge

What evaluating limits actually demands

Evaluating limits is the gateway skill of AP Calculus AB, and the no-calculator section leans on it heavily. The College Board wants you to find an exact value, show the algebra that justifies it, and recognize instantly which technique a given limit needs. This guide ties together the matching dot-point pages, each with its own practice: limits using algebraic properties, limits using algebraic manipulation, selecting procedures for limits, the squeeze theorem, infinite limits and vertical asymptotes, and limits at infinity and horizontal asymptotes.

The substitution-first routine

Every limit starts the same way: substitute the target value. The result tells you the problem type.

  • A real number means the function is continuous at aa, and that number is the limit. Done.
  • A nonzero number over zero (k0\frac{k}{0}, kβ‰ 0k \neq 0) means an infinite limit; the finite limit does not exist, and you do sign analysis to report +∞+\infty or βˆ’βˆž-\infty.
  • 00\frac{0}{0} is indeterminate: the limit may exist, but you must rewrite the function first.

This single routine prevents the most common mistake of all, which is reporting 00\frac{0}{0} as if it were a value.

Resolving 0/0 by rewriting

When substitution gives 00\frac{0}{0}, the numerator and denominator share a hidden common factor. Match the obstacle to the tool.

Factor and cancel

For rational expressions, factor both parts and cancel the factor that vanishes at aa:

lim⁑xβ†’4x2βˆ’16xβˆ’4=lim⁑xβ†’4(xβˆ’4)(x+4)xβˆ’4=lim⁑xβ†’4(x+4)=8.\lim_{x \to 4} \frac{x^2 - 16}{x - 4} = \lim_{x \to 4} \frac{(x-4)(x+4)}{x-4} = \lim_{x \to 4}(x + 4) = 8.

Multiply by the conjugate

When a square root creates the 00\frac{0}{0}, multiply top and bottom by the conjugate to convert the root into a difference of squares:

lim⁑xβ†’0x+1βˆ’1x=lim⁑xβ†’0xx(x+1+1)=lim⁑xβ†’01x+1+1=12.\lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x} = \lim_{x \to 0} \frac{x}{x(\sqrt{x+1}+1)} = \lim_{x \to 0} \frac{1}{\sqrt{x+1}+1} = \frac{1}{2}.

Combine a complex fraction

When fractions are stacked, combine them over a common denominator before cancelling:

lim⁑xβ†’31xβˆ’13xβˆ’3=lim⁑xβ†’33βˆ’x3xxβˆ’3=lim⁑xβ†’3βˆ’13x=βˆ’19.\lim_{x \to 3} \frac{\frac{1}{x} - \frac{1}{3}}{x - 3} = \lim_{x \to 3} \frac{\frac{3 - x}{3x}}{x - 3} = \lim_{x \to 3} \frac{-1}{3x} = -\frac{1}{9}.

The special trigonometric limits

Two limits (in radians) resolve most trigonometric indeterminate forms:

lim⁑xβ†’0sin⁑xx=1,lim⁑xβ†’01βˆ’cos⁑xx=0.\lim_{x \to 0} \frac{\sin x}{x} = 1, \qquad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0.

The trick is to rewrite until one pattern appears. For lim⁑xβ†’0sin⁑3xx\lim_{x \to 0} \frac{\sin 3x}{x}, multiply and divide to match sin⁑3x3x\frac{\sin 3x}{3x}:

lim⁑xβ†’0sin⁑3xx=lim⁑xβ†’03β‹…sin⁑3x3x=3(1)=3.\lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} 3 \cdot \frac{\sin 3x}{3x} = 3(1) = 3.

Infinite limits and sign analysis

When substitution gives k0\frac{k}{0} with kβ‰ 0k \neq 0, the line x=ax = a is a vertical asymptote and the limit is infinite. Determine the sign on each side: check the sign of the numerator near aa and the sign of the (tiny) denominator just left and just right of aa. Matching signs give +∞+\infty; opposite signs give βˆ’βˆž-\infty. A squared factor in the denominator gives the same infinity on both sides.

Limits at infinity

For end behavior as xβ†’Β±βˆžx \to \pm\infty, divide every term by the highest power of xx in the denominator and send the 1xn\frac{1}{x^n} terms to zero:

lim⁑xβ†’βˆž3x2βˆ’x6x2+5=lim⁑xβ†’βˆž3βˆ’1x6+5x2=36=12.\lim_{x \to \infty} \frac{3x^2 - x}{6x^2 + 5} = \lim_{x \to \infty} \frac{3 - \frac{1}{x}}{6 + \frac{5}{x^2}} = \frac{3}{6} = \frac{1}{2}.

The degree rule summarizes the outcome: bottom-heavy gives 00, equal degrees give the leading-coefficient ratio, and top-heavy gives ±∞\pm\infty. Watch the sign of x2=∣x∣\sqrt{x^2} = |x|, which equals βˆ’x-x as xβ†’βˆ’βˆžx \to -\infty and can produce two different horizontal asymptotes.

How limit evaluation is examined

  • Direct evaluation. Substitute and simplify; report exact values, not decimals, on the no-calculator part.
  • Indeterminate forms. Recognize 00\frac{0}{0} and choose factoring, a conjugate, fraction-combining, or a special trig limit.
  • Infinite behavior. Identify vertical asymptotes with sign analysis and horizontal asymptotes with the degree rule.
  • Connecting representations. Confirm an algebraic limit against a graph or table when the question gives one.

Check your knowledge

A mix of direct, indeterminate, infinite and at-infinity limits. Work them under no-calculator conditions, then check against the solutions.

  1. Evaluate lim⁑xβ†’1(2x2βˆ’3x+5)\lim_{x \to 1}(2x^2 - 3x + 5). (1 mark)
  2. Evaluate lim⁑xβ†’3x2βˆ’9xβˆ’3\lim_{x \to 3} \frac{x^2 - 9}{x - 3}. (2 marks)
  3. Evaluate lim⁑xβ†’0x+4βˆ’2x\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}. (2 marks)
  4. Evaluate lim⁑xβ†’0sin⁑4xx\lim_{x \to 0} \frac{\sin 4x}{x}. (2 marks)
  5. Evaluate lim⁑xβ†’2+1xβˆ’2\lim_{x \to 2^+} \frac{1}{x - 2} and state its sign. (2 marks)
  6. Evaluate lim⁑xβ†’βˆž5x2+23x2βˆ’x\lim_{x \to \infty} \frac{5x^2 + 2}{3x^2 - x}. (2 marks)
  7. Evaluate lim⁑xβ†’βˆ’βˆž2x+1x2+1\lim_{x \to -\infty} \frac{2x + 1}{\sqrt{x^2 + 1}}. (2 marks)
  8. Evaluate lim⁑xβ†’41xβˆ’14xβˆ’4\lim_{x \to 4} \frac{\frac{1}{x} - \frac{1}{4}}{x - 4}. (2 marks)

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  • calculus
  • ap
  • ap-calculus-ab
  • limits
  • evaluating-limits
  • indeterminate-form
  • exam-technique