Lecture Note 2: Limits and Continuity of Functions

1. Introductory Review: Functions and Composition

This lecture builds on the idea of a function as a rule that assigns every permitted input exactly one output. Before studying limits, students should be confident with function evaluation, piecewise functions, composite functions, domains, and ranges.

A piecewise function is a function whose rule changes depending on the interval in which the input lies. A composite function is formed when the output of one function becomes the input of another function.

The composite of f with g is written as:

(f ∘ g)(x) = f(g(x))

This means that g is applied first, and f is applied afterward.

Example 1

Given f(x) = √x and g(x) = x² - 1, evaluate f(g(1)) and g(f(1)).

Solution:

g(1) = 1² - 1 = 0

f(g(1)) = f(0) = √0 = 0

Also:

f(1) = √1 = 1

g(f(1)) = g(1) = 1² - 1 = 0

In this example, both results are equal. However, this does not always happen. In general:

f(g(x)) ≠ g(f(x))

Exam Points

For f(g(x)), begin with the inner function g(x).
The domain of f ∘ g contains only values of x for which g(x) is defined and g(x) is allowed inside f.
Composite functions are important because limits and continuity often depend on how functions are combined.

2. Limits of Functions

A limit describes the value that a function approaches as the input approaches a specified number.

Suppose a function is not defined at x = a, but it is defined for values of x close to a. The limit may still exist if the function values approach one definite number as x gets closer and closer to a.

Symbolically, if f(x) approaches L as x approaches a, we write:

lim_{x → a} f(x) = L

This is read as: "the limit of f(x) as x approaches a is L."

Illustrative Example

Consider:

f(x) = ^{x² - 4}⁄_{x - 2}

At x = 2, direct substitution gives:

^{2² - 4}⁄_{2 - 2} = ⁰⁄₀

This is undefined. However, we can simplify the function for x ≠ 2:

^{x² - 4}⁄_{x - 2} = ^{(x - 2)(x + 2)}⁄_{x - 2} = x + 2, for x ≠ 2

Therefore:

lim_{x → 2} ^{x² - 4}⁄_{x - 2} = lim_{x → 2} (x + 2) = 4

This shows that a limit may exist even when the function itself is not defined at that point.

Formal Idea of a Limit

The statement lim_{x → a} f(x) = L means that f(x) can be made as close as desired to L by taking x sufficiently close to a, but not necessarily equal to a.

In intuitive absolute-value language:

|f(x) - L| → 0 as |x - a| → 0

Key Takeaways

A limit studies the behavior of a function near a point, not necessarily at the point.
The value f(a) may be undefined even when lim_{x → a} f(x) exists.
If direct substitution gives a valid real number, the limit is usually obtained by substitution.
If direct substitution gives ⁰⁄₀, algebraic simplification is required.

3. Basic Limit Laws

Let lim_{x → a} f(x) = L and lim_{x → a} g(x) = K. The following laws are fundamental.

1. Constant Law

lim_{x → a} c = c

Example:

lim_{x → 2} 3 = 3

2. Identity Law

lim_{x → a} x = a

Example:

lim_{x → -4} x = -4

3. Power Law

lim_{x → a} xⁿ = aⁿ, where n is a positive integer.

Example:

lim_{x → 2} x² = 2² = 4

4. Constant Multiple Law

lim_{x → a} [c f(x)] = cL

Example:

lim_{x → 2} 3x² = 3(2²) = 12

5. Sum and Difference Laws

lim_{x → a} [f(x) ± g(x)] = L ± K

Example:

lim_{x → 5} (2x² + 4x) = 2(5²) + 4(5) = 70

6. Product Law

lim_{x → a} [f(x)g(x)] = LK

Example:

lim_{x → 1} (x² + 1)(4x + 13) = (2)(17) = 34

7. Quotient Law

lim_{x → a} ^f(x)⁄_g(x) = ^L⁄_K, provided K ≠ 0.

Example:

lim_{x → -2} ^{x² + 2}⁄_{2 - x} = ⁶⁄₄ = ³⁄₂

8. Root Law

For suitable values in the domain:

lim_{x → a} ⁿ√f(x) = ⁿ√L

For even roots, the expression under the root must be non-negative near the point of approach.

9. Polynomial Law

If P(x) is a polynomial, then:

lim_{x → a} P(x) = P(a)

Example:

lim_{x → 3} (2x³ + 6x² - 3x + 1) = 100

Exam Points

Polynomial limits are evaluated by direct substitution.
Rational-function limits are evaluated by substitution only if the denominator is not zero.
The quotient law requires the limiting denominator to be non-zero.
Limit laws allow complicated limits to be broken into simpler parts.

4. Evaluating Limits by Algebraic Simplification

When direct substitution gives ⁰⁄₀, the expression is called an indeterminate form. This does not mean the limit is zero or undefined. It means further work is needed.

The two most common algebraic techniques are:

Factorization and cancellation.
Rationalization using the conjugate.

Example 2: Factorization

Evaluate:

lim_{x → -3} ^{x² + x - 6}⁄_{x + 3}

Solution:

Direct substitution gives ⁰⁄₀. Factor the numerator:

x² + x - 6 = (x + 3)(x - 2)

Therefore:

lim_{x → -3} ^{x² + x - 6}⁄_{x + 3}

= lim_{x → -3} ^{(x + 3)(x - 2)}⁄_{x + 3}

= lim_{x → -3} (x - 2)

= -3 - 2

= -5

Example 3: Factorization in Numerator and Denominator

Evaluate:

lim_{x → 2} ^{x² - 3x + 2}⁄_{x² - 6x + 8}

Solution:

Factor both numerator and denominator:

x² - 3x + 2 = (x - 2)(x - 1)

x² - 6x + 8 = (x - 2)(x - 4)

Thus:

lim_{x → 2} ^{(x - 2)(x - 1)}⁄_{(x - 2)(x - 4)}

= lim_{x → 2} ^{x - 1}⁄_{x - 4}

= ^{2 - 1}⁄_{2 - 4}

= -¹⁄₂

Example 4: Rationalization

Evaluate:

lim_{x → 0} ^{√(x² + 9) - 3}⁄_x²

Solution:

Multiply by the conjugate:

^{√(x² + 9) - 3}⁄_x² × ^{√(x² + 9) + 3}⁄_{√(x² + 9) + 3}

= ^{(x² + 9) - 9}⁄_{x²[√(x² + 9) + 3]}

= ^x²⁄_{x²[√(x² + 9) + 3]}

= ¹⁄_{√(x² + 9) + 3}

Now substitute x = 0:

= ¹⁄_{√9 + 3}

= ¹⁄₆

Example 5: Rationalization with Two Radicals

Evaluate:

lim_{x → 1} ^{√(5x - 4) - √x}⁄_{x - 1}

Solution:

Multiply by the conjugate:

^{√(5x - 4) - √x}⁄_{x - 1} × ^{√(5x - 4) + √x}⁄_{√(5x - 4) + √x}

= ^{(5x - 4) - x}⁄_{(x - 1)[√(5x - 4) + √x]}

= ^{4x - 4}⁄_{(x - 1)[√(5x - 4) + √x]}

= ^{4(x - 1)}⁄_{(x - 1)[√(5x - 4) + √x]}

= ⁴⁄_{√(5x - 4) + √x}

Now substitute x = 1:

= ⁴⁄_{√1 + √1}

= 2

Key Takeaways

The form ⁰⁄₀ is indeterminate, not automatically zero.
Factorization helps remove a common zero factor.
Rationalization is useful when a radical expression causes ⁰⁄₀.
After simplification, substitute the limiting value.

5. Infinite Limits

An infinite limit occurs when the values of a function increase or decrease without bound as x approaches a particular number.

If f(x) becomes arbitrarily large as x approaches a, we write:

lim_{x → a} f(x) = ∞

If f(x) becomes arbitrarily negative as x approaches a, we write:

lim_{x → a} f(x) = -∞

Strictly speaking, ∞ and -∞ are not real-number limits. They describe unbounded behavior.

One-Sided Behavior

Consider:

f(x) = ³⁄_{x - 2}

As x approaches 2 from the right, x - 2 is positive and very small, so f(x) becomes very large:

lim_{x → 2⁺} ³⁄_{x - 2} = ∞

As x approaches 2 from the left, x - 2 is negative and very small, so f(x) becomes very negative:

lim_{x → 2^-} ³⁄_{x - 2} = -∞

Therefore, the two-sided limit does not exist as a single infinite direction.

Example 6

Evaluate:

lim_{x → 3} ^-2⁄_{(x - 3)²}

Solution:

As x approaches 3, (x - 3)² approaches 0 through positive values. Since the numerator is negative, the quotient decreases without bound.

Therefore:

lim_{x → 3} ^-2⁄_{(x - 3)²} = -∞

Key Takeaways

Infinite limits describe unbounded behavior.
A vertical asymptote often occurs where a denominator approaches zero.
Always check left-hand and right-hand behavior when signs may change.
If one side approaches ∞ and the other approaches -∞, the two-sided infinite behavior is not the same.

6. Limits as x Approaches Infinity

A limit as x → ∞ describes the long-run or end behavior of a function.

For a positive integer n:

lim_{x → ∞} ¹⁄_xⁿ = 0

More generally, for any constant a:

lim_{x → ∞} ^a⁄_xⁿ = 0

Rational Functions at Infinity

For rational functions, compare the highest powers of x in the numerator and denominator.

If the degree of the numerator is less than the degree of the denominator, the limit is 0.
If the degrees are equal, the limit is the ratio of the leading coefficients.
If the degree of the numerator is greater than the degree of the denominator, the magnitude grows without bound, unless signs or directions require special analysis.

Example 7

Evaluate:

lim_{x → ∞} ^{x² + 6x + 2}⁄_{x³ + x - 13}

Solution:

The numerator has degree 2, while the denominator has degree 3. Since the denominator has the higher degree:

lim_{x → ∞} ^{x² + 6x + 2}⁄_{x³ + x - 13} = 0

Example 8

Evaluate:

lim_{x → ∞} ^{2x² - 6x + 1}⁄_{4x² + x - 3}

Solution:

The numerator and denominator both have degree 2. Therefore, the limit is the ratio of the leading coefficients:

²⁄₄ = ¹⁄₂

Thus:

lim_{x → ∞} ^{2x² - 6x + 1}⁄_{4x² + x - 3} = ¹⁄₂

Example 9

Evaluate:

lim_{x → ∞} ^{x³ + 2x² + 7}⁄_{4x² + 3x}

Solution:

The numerator has degree 3, while the denominator has degree 2. The numerator grows faster than the denominator. Since the leading terms are positive:

lim_{x → ∞} ^{x³ + 2x² + 7}⁄_{4x² + 3x} = ∞

Exam Points

At infinity, the highest powers dominate.
Lower-degree terms become insignificant compared with the leading term.
For rational functions, compare degrees first.
When degrees are equal, use the ratio of leading coefficients.

7. Continuity of a Function

A function is continuous at a point if its graph has no break, hole, or jump at that point.

Formally, a function f is continuous at x = a if all three conditions hold:

f(a) is defined.
lim_{x → a} f(x) exists.
lim_{x → a} f(x) = f(a).

If any of these three conditions fails, the function is discontinuous at x = a.

Example 10

Determine whether f(x) = x² + 6x is continuous at x = -2.

Solution:

First:

f(-2) = (-2)² + 6(-2) = 4 - 12 = -8

So f(-2) is defined.

lim_{x → -2} (x² + 6x) = (-2)² + 6(-2) = -8

Therefore:

lim_{x → -2} f(x) = f(-2)

Hence, f is continuous at x = -2.

Example 11

Determine whether g(x) = ¹⁄_x² is continuous at x = 0.

Solution:

At x = 0:

g(0) = ¹⁄_0²

This is undefined. Since the first condition for continuity fails, g is not continuous at x = 0.

Example 12

Determine whether h(x) = cos(2x) is continuous at x = ^π⁄₂.

Solution:

h(^π⁄₂) = cos(2 × ^π⁄₂) = cos π = -1

Also:

lim_{x → π/2} cos(2x) = cos π = -1

Therefore:

lim_{x → π/2} h(x) = h(^π⁄₂)

So h is continuous at x = ^π⁄₂.

Key Takeaways

Continuity requires the function value, the limit, and equality between them.
Polynomials are continuous for all real numbers.
Rational functions are continuous wherever their denominators are not zero.
Trigonometric functions such as sin x and cos x are continuous for all real numbers.

8. Types of Discontinuity

A discontinuity occurs where a function fails to be continuous.

1. Removable Discontinuity

A removable discontinuity occurs when the limit exists at a point but the function is either undefined there or assigned the wrong value. It appears as a hole in the graph.

Example 13

Consider:

f(x) = ^{x² - x - 2}⁄_{x - 2}

Factor the numerator:

x² - x - 2 = (x - 2)(x + 1)

Thus:

f(x) = x + 1, for x ≠ 2

The function is not defined at x = 2, but:

lim_{x → 2} f(x) = 3

Therefore, f has a removable discontinuity at x = 2. If we define f(2) = 3, the discontinuity is removed.

2. Infinite Discontinuity

An infinite discontinuity occurs when the function grows without bound near a point. It is usually associated with a vertical asymptote.

Example:

f(x) = ¹⁄_{(x - 1)²}

The function is discontinuous at x = 1 because the denominator becomes zero and the function increases without bound.

3. Jump Discontinuity

A jump discontinuity occurs when the left-hand and right-hand limits exist but are not equal. This often appears in piecewise-defined functions.

Exam Points

A removable discontinuity can be repaired by redefining the function at one point.
An infinite discontinuity is associated with unbounded behavior.
A jump discontinuity occurs when the two one-sided limits are different.
To test continuity, always check the three formal conditions.

9. Continuity on an Interval

A function is continuous on an interval if it is continuous at every point in that interval.

Example 14

Discuss the continuity of:

f(x) = ^{x² - 1}⁄_{x - 1}

Solution:

The function is undefined at x = 1. Factor the numerator:

x² - 1 = (x - 1)(x + 1)

Thus, for x ≠ 1:

f(x) = x + 1

Therefore:

lim_{x → 1} f(x) = 2

The function has a removable discontinuity at x = 1. If we define f(1) = 2, the new function becomes continuous for all real numbers.

Example 15

Discuss the continuity of:

g(x) = x + 1, if x ≤ 0

g(x) = x² + 1, if x > 0

Solution:

For x < 0, g(x) = x + 1, a polynomial, so it is continuous.

For x > 0, g(x) = x² + 1, also a polynomial, so it is continuous.

At x = 0:

g(0) = 0 + 1 = 1

Left-hand limit:

lim_{x → 0^-} g(x) = 1

Right-hand limit:

lim_{x → 0⁺} g(x) = 0² + 1 = 1

Since both one-sided limits equal g(0), the function is continuous at x = 0. Therefore, g is continuous for all real numbers.

Key Takeaways

To discuss continuity on an interval, identify possible trouble points first.
For rational functions, trouble points occur where the denominator is zero.
For piecewise functions, check the boundary points where the rule changes.
If a discontinuity can be fixed by defining one missing value, it is removable.

10. Exam Formula and Concept Summary

Limit Notation

lim_{x → a} f(x) = L

Intuitive Meaning

|f(x) - L| → 0 as |x - a| → 0

Direct Substitution

If f is continuous at a, then:

lim_{x → a} f(x) = f(a)

Polynomial Limit

lim_{x → a} P(x) = P(a)

Quotient Law

lim_{x → a} ^f(x)⁄_g(x) = ^L⁄_K, provided K ≠ 0.

Limit at Infinity

lim_{x → ∞} ^a⁄_xⁿ = 0, for n > 0.

Continuity at a Point

A function f is continuous at x = a if:

f(a) is defined.
lim_{x → a} f(x) exists.
lim_{x → a} f(x) = f(a).

11. Practice Questions for Revision

A. Basic Limits

Evaluate lim_{x → -3} (2x² + 4x + 1).
Evaluate lim_{x → 1} (3x³ - 2x² + 4).
Evaluate lim_{x → -4} (x + 3)².
Evaluate lim_{x → 0} (2x - 1)³.
Evaluate lim_{x → 4} ³√(x + 4).

B. Algebraic Limits

Evaluate lim_{h → 0} ^{(3 + h)² - 9}⁄_h.
Evaluate lim_{x → -4} ^{x² + 5x + 4}⁄_{x² + 3x - 4}.
Evaluate lim_{x → -3} ^{x² - 9}⁄_{2x² + 7x + 3}.
Evaluate lim_{x → -4} ^{√(x² + 9) - 5}⁄_{x + 4}.
Evaluate lim_{x → 0} ^{√(1 + x) - 1}⁄_x.

C. Limits at Infinity

Evaluate lim_{x → ∞} ^{2x³ + 6x² + 4x}⁄_{6x³ + x² + 3x}.
Evaluate lim_{x → ∞} ^{16x⁴ + 13x² + 2x}⁄_{5x³ + 16x² + 2}.
Evaluate lim_{x → ∞} ^{x⁴ + 6x³ + 21x}⁄_{6x⁵ + 3x⁴ + 6x²}.

D. Continuity

Discuss the continuity of f(x) = ¹⁄_{(x - 1)²}.
Discuss the continuity of h(x) = ^x⁄_{x² - x}.
Discuss the continuity of g(x) = ¹⁄_{x - 1}, if x ≠ 1, and g(1) = 2.
Discuss the continuity of F(x) = ^{x² - x}⁄_{x - 1}, if x ≠ 1, and F(1) = 1.
Discuss the continuity of F(x) = ^{2x² - 5x - 3}⁄_{x - 3}, if x ≠ 3, and F(3) = 6.

12. Final Revision Checklist

Before an examination, make sure you can:

Explain the meaning of a limit in words and symbols.
Evaluate simple limits by direct substitution.
Recognize the indeterminate form ⁰⁄₀.
Use factorization to evaluate limits.
Use rationalization to evaluate limits containing radicals.
Evaluate limits at infinity using leading powers.
Distinguish between finite limits and infinite limits.
Apply the three conditions for continuity at a point.
Identify removable, infinite, and jump discontinuities.
Discuss continuity on an interval.