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:

limx → 2+ 3⁄x - 2 = ∞

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

limx → 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:

1. f(a) is defined.
2. lim_{x → a} f(x) exists.
3. 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.

Next:

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:

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

11. Practice Questions for Revision

A. Basic Limits

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

B. Algebraic Limits

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

C. Limits at Infinity

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

D. Continuity

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