Module 1: Limits and Continuity

Calculus I – Study Guide

Learn Without Walls

1. Limits

Informal Definition

limx→a f(x) = L means f(x) gets arbitrarily close to L as x approaches a (from both sides), regardless of the value of f(a).

One-Sided Limits

Limit Laws

LawStatement
Sumlim [f + g] = lim f + lim g
Differencelim [f − g] = lim f − lim g
Productlim [f · g] = (lim f)(lim g)
Quotientlim [f/g] = (lim f)/(lim g), g ≠ 0
Powerlim [f]n = (lim f)n
Constant Multiplelim [cf] = c · lim f

Algebraic Techniques

  1. Direct substitution — try first; works if no 0/0.
  2. Factor and cancel — factor numerator/denominator, cancel common factor.
  3. Rationalize — multiply by conjugate to clear radicals.
  4. Trig identities — use (sin x)/x → 1 and (1 − cos x)/x → 0.
  5. Squeeze Theorem — trap f between two functions with the same limit.
Special Limits:
limx→0 (sin x)/x = 1
limx→0 (1 − cos x)/x = 0

2. Limits at Infinity

Rational Functions at Infinity

Compare degrees of numerator (n) and denominator (m):
n < m: limit = 0 | n = m: limit = an/bm | n > m: limit = ±∞

Asymptotes

TypeHow to Find
Horizontal (y = L)limx→±∞ f(x) = L
Vertical (x = a)Denominator = 0 while numerator ≠ 0

3. Continuity

Three Conditions

f is continuous at x = a if: (1) f(a) is defined, (2) limx→a f(x) exists, (3) limx→a f(x) = f(a).

Types of Discontinuities

TypeDescription
RemovableLimit exists but f(a) is wrong or missing (hole)
JumpOne-sided limits both exist but differ
InfiniteAt least one side → ±∞ (vertical asymptote)

Families of Continuous Functions

Polynomials, rational functions (on domain), trig functions, exponentials, logarithms, and compositions/sums/products of these are continuous on their domains.

4. Intermediate Value Theorem

If f is continuous on [a, b] and N is between f(a) and f(b), then there exists c in (a, b) with f(c) = N.

Using the IVT

  1. Verify f is continuous on [a, b].
  2. Compute f(a) and f(b).
  3. Show that the target value N lies between f(a) and f(b).
  4. Conclude that a solution exists in (a, b).
Example: Show x³ − x − 1 = 0 has a root in [1, 2].
f(1) = −1 < 0, f(2) = 5 > 0. Continuous polynomial, sign change ⇒ root exists.