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Calculus I (Differential & intro Integral) Unit 1: Limits & continuity

A limit tells you the value f(x) heads toward as x gets close to a number, whether or not f actually equals that value there. Continuity is the case where the limit and the function value agree.

Covers: Limit notation, one-sided limits, limit laws, squeeze theorem, limits at infinity, continuity, IVT

Study guide

A limit describes the value a function approaches as the input approaches a number. The notation lim x→a f(x) = L means f(x) gets arbitrarily close to L as x gets close to a, regardless of what happens exactly at a. One-sided limits approach from the left (x→a⁻) or the right (x→a⁺), and the two-sided limit exists only when both one-sided limits exist and are equal.

Limit laws let you split limits over sums, products and quotients, and for continuous functions you can just substitute. When substitution gives 0/0, an indeterminate form, simplify first: factor and cancel, multiply by a conjugate to rationalize, or use an identity. The squeeze theorem handles functions trapped between two others with the same limit; it proves lim x→0 (sin x)/x = 1.

Limits at infinity describe end behavior and horizontal asymptotes. For rational functions, compare degrees: a bigger denominator degree gives 0, equal degrees give the ratio of leading coefficients, and a bigger numerator degree means no finite limit. Infinite limits signal vertical asymptotes.

A function is continuous at a if f(a) is defined, the limit exists, and they're equal. Discontinuities are removable (holes), jumps, or infinite. The Intermediate Value Theorem says a continuous function on [a, b] takes every value between f(a) and f(b), which guarantees solutions exist.

As x → ∞, e^(−x) → 0 while eˣ → ∞, and arctan x → π/2 (with arctan x → −π/2 as x → −∞), so a model such as a + b·arctan t levels off at a + bπ/2. The function ln x grows without bound, but more slowly than any positive power of x. To justify an answer with the IVT, state that f is continuous on [a, b] (differentiable functions are continuous), give f(a) and f(b), and note that the target value lies between them. A table of values alone can't guarantee continuity unless the problem says f is continuous or differentiable.

To make a piecewise function continuous at a break point c, evaluate the left piece at c and the right piece at c, set the two results equal (and equal to f(c)), and solve for the unknown constant. Related limits follow from lim x→0 (sin x)/x = 1: lim x→0 sin(kx)/x = k, and lim x→0 (1 − cos x)/x = 0.

Key ideas

Formulas and rules

lim x→0 (sin x)/x = 1
lim x→0 (1 − cos x)/x = 0
Continuous at a ⇔ f(a) defined, lim f(x) exists, and they are equal

Common mistakes

Words to know

Two-sided limit
Exists when the left- and right-hand limits exist and are equal.
Continuity at x = a
f(a) is defined, the limit exists, and the limit equals f(a).
lim x→0 (sin x)/x
1
Indeterminate form 0/0
Simplify (factor, rationalize, identity), then evaluate again.
Intermediate Value Theorem
If f is continuous on [a, b] and N is between f(a) and f(b), then f(c) = N for some c in (a, b).
Limit at infinity (equal degrees)
The ratio of the leading coefficients: the horizontal asymptote.
Squeeze theorem
If g ≤ f ≤ h and g, h → L, then f → L.
Limit
The value f(x) approaches as x approaches a.
One-sided limit
The limit as x approaches a from the left (a⁻) or right (a⁺).
Limit laws
Limits split over sums, products and quotients (denominator limit ≠ 0).
Infinite limit
f(x) grows without bound near a: a vertical asymptote.
Limit at infinity
End behavior as x → ±∞: a horizontal asymptote if finite.
Removable discontinuity
A hole: the limit exists but f(a) is missing or different.
Jump discontinuity
One-sided limits exist but aren't equal.
Rationalizing (limits)
Multiply by a conjugate to resolve a 0/0 form with roots.

Sample practice questions

49 more questions like these, from level 1 to 5, are in the practice set.

Question 1 (level 1)

lim x→2 (3x + 1)

  1. 5
  2. Does not exist
  3. 7
  4. 6
Show the answer and steps

Answer: C) 7

  1. 3x + 1 is a polynomial, so it's continuous.
  2. Plug in: 3(2) + 1 = 7.
Question 2 (level 1)

Find lim x→−2 (2x² − 2).

  1. Does not exist
  2. 6
  3. 7
  4. −2
Show the answer and steps

Answer: B) 6

  1. Polynomials are continuous, so substitute directly.
  2. 2(−2)² + 0(−2) + (−2) = 6.
Question 3 (level 3)

lim x→∞ (2x² + 1)/(5x² − x)

  1. 0
  2. 2/5
  3. ∞
  4. 2
Show the answer and steps

Answer: B) 2/5

  1. Divide top and bottom by x²: (2 + 1/x²)/(5 − 1/x).
  2. As x → ∞ the small terms → 0.
  3. 2/5.