AP Calculus AB › Unit 1
Calculus rests on limits: the value a function approaches, which may differ from its value at the point. Unit 1 covers limits from graphs, tables, and algebra, the squeeze theorem, types of discontinuity, asymptotes, and the Intermediate Value Theorem. The exam asks you to evaluate or estimate limits in any representation and to justify continuity using the definition. A frequent mistake is assuming the limit always equals f(a). Another is applying the IVT without first stating that its conditions, such as continuity on a closed interval, are met.
These are taken from the Regular set, with the answer and explanation shown.
1. Let f(x) = x² − 4. If lim x→1 g(x) = 3, what is lim x→1 f(g(x))?
Answer: C. f is a polynomial, so it is continuous. That lets the limit pass inside: lim f(g(x)) = f(lim g(x)) = f(3) = 9 − 4 = 5.
Common trap: Choosing −3 computes f(1). The inside function g is near 3, not 1, so f must be evaluated at 3.
2. What is lim x→0 (sin x)/(x + tan x)?
Answer: B. Divide top and bottom by x: (sin x/x)/(1 + tan x/x). As x→0, sin x/x→1 and tan x/x→1. The limit is 1/(1 + 1) = 1/2.
Common trap: Choosing 1 treats the denominator like x alone. The tan x term adds about another x, so the bottom is about 2x.
3. What is lim x→−∞ √(9x² + 1)/(x − 2)?
Answer: B. For large |x|, √(9x² + 1) ≈ √(9x²) = 3|x|. When x is negative, |x| = −x. So the quotient ≈ 3(−x)/x = −3.
Common trap: Choosing 3 treats √(x²) as x. For x→−∞ the root is positive while x − 2 is negative, so the quotient is negative.