Limits & Continuity 1
LIMITS & CONTINUITY
The Foundation of Calculus
Complete Guide with 100+ Practice Problems
Everything You Need to Know About Limits:
Limit Definition • Limit Rules • One-Sided Limits •
Limits at Infinity
L’Hôpital’s Rule • Continuity • Intermediate Value
Theorem
y
y=2
As x → 1, f (x) → 0.5
(1, 0.5) x
Author: Rachid Ousalem
March 28, 2026
Rachid Ousalem
,Limits & Continuity 2
Contents
1 Introduction to Limits 3
2 One-Sided Limits 5
3 Practice Problems 7
Solutions to Basic Limits 8
4 Limits at Infinity 9
5 Indeterminate Forms and L’Hôpital’s Rule 14
6 Continuity 19
7 Intermediate Value Theorem (IVT) 23
Additional Limit Problems 27
8 Detailed Continuity Examples 29
9 More IVT Applications 31
10 Limits Involving Absolute Value 32
11 Squeeze Theorem 34
Quick Reference Card 39
Final Mastery Check 40
Formula Summary 42
Mega Practice: 100 Limit Problems 44
Solutions to Mega Practice 49
Challenge Problems 54
Quick Review Table 56
Quick Reference Card 58
Final 10 Quick Problems 59
Rachid Ousalem
,Limits & Continuity 3
1 Introduction to Limits
What is a Limit?
The limit of a function f (x) as x approaches a is the value that f (x) gets closer
and closer to as x gets arbitrarily close to a.
[Formula] Formula:Notation:
lim f (x) = L
x→a
[Tip] Pro Tip:The limit describes the behavior of f (x) near x = a, not necessarily
the value at x = a.
y
As x → 2, f (x) → 0.5
(2, 0.5) x
Rachid Ousalem
, Limits & Continuity 4
Example 1: Finding a Limit by Direct Substitution
Find lim (3x + 1).
x→2
Solution: Since f (x) = 3x + 1 is a polynomial (continuous everywhere), we can
substitute x = 2:
lim (3x + 1) = 3(2) + 1 = 7
x→2
[Tip] Pro Tip:For polynomials, rational functions (where denominator ̸= 0), and
other continuous functions, you can find the limit by direct substitution.
Example 2: When Direct Substitution Fails
x2 − 1
Find lim .
x→1 x − 1
Solution: Direct substitution gives 0
0
(indeterminate form). We need to simplify:
x2 − 1 (x − 1)(x + 1)
= = x + 1 (x ̸= 1)
x−1 x−1
Therefore:
x2 − 1
lim = lim (x + 1) = 2
x→1 x − 1 x→1
[Warning] Warning:The function is undefined at x = 1, but the limit exists and
equals 2!
Rachid Ousalem
LIMITS & CONTINUITY
The Foundation of Calculus
Complete Guide with 100+ Practice Problems
Everything You Need to Know About Limits:
Limit Definition • Limit Rules • One-Sided Limits •
Limits at Infinity
L’Hôpital’s Rule • Continuity • Intermediate Value
Theorem
y
y=2
As x → 1, f (x) → 0.5
(1, 0.5) x
Author: Rachid Ousalem
March 28, 2026
Rachid Ousalem
,Limits & Continuity 2
Contents
1 Introduction to Limits 3
2 One-Sided Limits 5
3 Practice Problems 7
Solutions to Basic Limits 8
4 Limits at Infinity 9
5 Indeterminate Forms and L’Hôpital’s Rule 14
6 Continuity 19
7 Intermediate Value Theorem (IVT) 23
Additional Limit Problems 27
8 Detailed Continuity Examples 29
9 More IVT Applications 31
10 Limits Involving Absolute Value 32
11 Squeeze Theorem 34
Quick Reference Card 39
Final Mastery Check 40
Formula Summary 42
Mega Practice: 100 Limit Problems 44
Solutions to Mega Practice 49
Challenge Problems 54
Quick Review Table 56
Quick Reference Card 58
Final 10 Quick Problems 59
Rachid Ousalem
,Limits & Continuity 3
1 Introduction to Limits
What is a Limit?
The limit of a function f (x) as x approaches a is the value that f (x) gets closer
and closer to as x gets arbitrarily close to a.
[Formula] Formula:Notation:
lim f (x) = L
x→a
[Tip] Pro Tip:The limit describes the behavior of f (x) near x = a, not necessarily
the value at x = a.
y
As x → 2, f (x) → 0.5
(2, 0.5) x
Rachid Ousalem
, Limits & Continuity 4
Example 1: Finding a Limit by Direct Substitution
Find lim (3x + 1).
x→2
Solution: Since f (x) = 3x + 1 is a polynomial (continuous everywhere), we can
substitute x = 2:
lim (3x + 1) = 3(2) + 1 = 7
x→2
[Tip] Pro Tip:For polynomials, rational functions (where denominator ̸= 0), and
other continuous functions, you can find the limit by direct substitution.
Example 2: When Direct Substitution Fails
x2 − 1
Find lim .
x→1 x − 1
Solution: Direct substitution gives 0
0
(indeterminate form). We need to simplify:
x2 − 1 (x − 1)(x + 1)
= = x + 1 (x ̸= 1)
x−1 x−1
Therefore:
x2 − 1
lim = lim (x + 1) = 2
x→1 x − 1 x→1
[Warning] Warning:The function is undefined at x = 1, but the limit exists and
equals 2!
Rachid Ousalem
