CALC-3.2

3-4 hours

Properties and Operations of Function Limits

Master the fundamental properties that make limits computable.

Learning Objectives

Prove and apply the uniqueness of function limits
Understand and use local boundedness of limits
Apply sign preservation theorems in proofs
Master all arithmetic operations on limits
Use the squeeze theorem to evaluate difficult limits
Apply the Cauchy criterion for function limits

1. Uniqueness of Function Limits

The first fundamental property: if a limit exists, it is unique. A function cannot approach two different values at the same point.

Theorem 3.4: Uniqueness of Function Limits

If $\lim_{x \to x_0} f(x) = L_1$ and $\lim_{x \to x_0} f(x) = L_2$ , then $L_1 = L_2$ .

Proof of Theorem 3.4:

Suppose $L_1 \neq L_2$ . Let $\varepsilon = |L_1 - L_2|/2 > 0$ .

By definitions, $\exists \delta_1, \delta_2 > 0$ such that:

0 < |x - x_0| < \delta_i \Rightarrow |f(x) - L_i| < \varepsilon

Let $\delta = \min(\delta_1, \delta_2)$ . For $0 < |x - x_0| < \delta$ :

|L_1 - L_2| \leq |f(x) - L_1| + |f(x) - L_2| < 2\varepsilon = |L_1 - L_2|

Contradiction! So $L_1 = L_2$ . ∎

∎

Remark 3.4: Importance

Uniqueness lets us write "THE limit" and use limits in equations unambiguously.

2. Local Boundedness

If a limit exists, the function cannot "blow up" near the limit point.

Theorem 3.5: Local Boundedness

If $\lim_{x \to x_0} f(x) = L$ , then f is locally bounded:

\exists \delta > 0, M > 0: 0 < |x - x_0| < \delta \Rightarrow |f(x)| \leq M

Proof of Theorem 3.5:

Take $\varepsilon = 1$ . $\exists \delta > 0$ : $0 < |x - x_0| < \delta \Rightarrow |f(x) - L| < 1$ .

By triangle inequality: $|f(x)| \leq |f(x) - L| + |L| < 1 + |L|$ .

So $M = 1 + |L|$ works. ∎

∎

Remark 3.5: Local ≠ Global

$f(x) = 1/x$ has limit 1 at x = 1, but is unbounded on (0, ∞).

Corollary 3.2: Bounded × Zero = Zero

If $\lim f(x) = 0$ and g is bounded near $x_0$ , then $\lim f(x)g(x) = 0$ .

Example 3.6: Application

If $\lim_{x \to 0} f(x) = 3$ , then $|f(x)| < 4$ near 0.

By local boundedness: $|f(x)| < 1 + |3| = 4$ in some U'(0, δ).

3. Sign Preservation

If a limit is positive (or negative), the function has the same sign near the limit point.

Theorem 3.6: Sign Preservation (Limit → Function)

If $\lim_{x \to x_0} f(x) = L > 0$ , then $\exists \delta > 0$ :

0 < |x - x_0| < \delta \Rightarrow f(x) > 0

Proof of Theorem 3.6:

Take $\varepsilon = L/2 > 0$ . Then $\exists \delta$ : $|f(x) - L| < L/2$ .

So $L/2 < f(x) < 3L/2$ . Since L > 0: $f(x) > L/2 > 0$ . ∎

∎

Theorem 3.7: Sign Preservation (Function → Limit)

If $f(x) \geq 0$ near $x_0$ and limit exists, then $\lim f(x) \geq 0$ .

Warning: Strict Inequality NOT Preserved

f(x) < g(x) does NOT imply lim f < lim g.

Example: 0 < x² for x ≠ 0, but lim 0 = lim x² = 0.

Theorem 3.8: Order Preservation

If $f(x) \leq g(x)$ near $x_0$ and both limits exist, then $\lim f \leq \lim g$ .

Example 3.7: Using Sign Preservation

If $\lim_{x \to 2} f(x) = 5$ , show f(x) > 3 near x = 2.

Let g(x) = f(x) - 3. Then lim g = 5 - 3 = 2 > 0.

By sign preservation: g(x) > 0 near 2, so f(x) > 3. ∎

4. Arithmetic Operations on Limits

Theorem 3.9: Limit Laws

If $\lim f(x) = a$ and $\lim g(x) = b$ , then:

\lim [f(x) \pm g(x)] = a \pm b

\lim [f(x) \cdot g(x)] = a \cdot b

\lim \frac{f(x)}{g(x)} = \frac{a}{b} \quad (b \neq 0)

\lim [c \cdot f(x)] = c \cdot a

\lim [f(x)]^n = a^n

Proof of Sum Rule:

Given ε > 0, find δ₁, δ₂ such that |f(x) - a| < ε/2 and |g(x) - b| < ε/2.

For δ = min(δ₁, δ₂):

|f(x) + g(x) - (a+b)| \leq |f(x)-a| + |g(x)-b| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon

∎

Example 3.8: Direct Application

Find $\lim_{x \to 3} (x^2 - 2x + 4)$ .

\lim_{x \to 3} (x^2 - 2x + 4) = 9 - 6 + 4 = 7

Example 3.9: Factor First

Find $\lim_{x \to 2} \frac{x^2 + 3x - 10}{x - 2}$ .

Factor: $x^2 + 3x - 10 = (x-2)(x+5)$

\lim_{x \to 2} \frac{(x-2)(x+5)}{x-2} = \lim_{x \to 2} (x+5) = 7

Example 3.10: Rationalize

Find $\lim_{x \to 4} \frac{x - 4}{\sqrt{x} - 2}$ .

\frac{x-4}{\sqrt{x}-2} \cdot \frac{\sqrt{x}+2}{\sqrt{x}+2} = \sqrt{x} + 2 \to 4

Corollary 3.3: Polynomial Limits

For polynomial P(x): $\lim_{x \to c} P(x) = P(c)$ .

Corollary 3.4: Rational Limits

For R(x) = P(x)/Q(x) with Q(c) ≠ 0: $\lim_{x \to c} R(x) = R(c)$ .

5. The Squeeze Theorem

The Squeeze Theorem is powerful for limits involving oscillating functions or products.

Theorem 3.10: Squeeze Theorem

If $f(x) \leq g(x) \leq h(x)$ near $x_0$ and $\lim f(x) = \lim h(x) = L$ , then:

\lim_{x \to x_0} g(x) = L

Proof of Theorem 3.10:

Given ε > 0, find δ such that L - ε < f(x) and h(x) < L + ε near x₀.

Since f ≤ g ≤ h:

L - \varepsilon < f(x) \leq g(x) \leq h(x) < L + \varepsilon

So |g(x) - L| < ε. ∎

∎

When to Use Squeeze

Functions with sin, cos (bounded oscillation)
Products: (→ 0) × (bounded)
When direct methods fail

Example 3.11: Classic: x·sin(1/x)

Find $\lim_{x \to 0} x \sin(1/x)$ .

Since $-1 \leq \sin(1/x) \leq 1$ :

-|x| \leq x \sin(1/x) \leq |x|

Both bounds → 0. By Squeeze: limit = 0.

Example 3.12: x²·cos(1/x²)

Find $\lim_{x \to 0} x^2 \cos(1/x^2)$ .

-x^2 \leq x^2 \cos(1/x^2) \leq x^2 \to 0

Example 3.13: At Infinity

Find $\lim_{x \to +\infty} \frac{\sin x}{x}$ .

\frac{-1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} \to 0

6. Cauchy Criterion

The Cauchy criterion characterizes limit existence: values get arbitrarily close to EACH OTHER near the limit point.

Theorem 3.11: Cauchy Criterion

$\lim_{x \to x_0} f(x)$ exists iff:

\forall \varepsilon > 0, \exists \delta > 0: x', x'' \in U'(x_0, \delta) \Rightarrow |f(x') - f(x'')| < \varepsilon

Proof of (⇒):

If lim f = L, find δ: |f(x) - L| < ε/2 for x ∈ U'(x₀, δ).

Then |f(x') - f(x'')| ≤ |f(x') - L| + |f(x'') - L| < ε. ∎

∎

Theorem 3.12: Cauchy at Infinity

$\lim_{x \to +\infty} f(x)$ exists iff:

\forall \varepsilon > 0, \exists M > 0: x', x'' > M \Rightarrow |f(x') - f(x'')| < \varepsilon

Example 3.14: Disproving with Cauchy

Show $\lim_{x \to +\infty} \cos(x)$ does not exist.

Take ε₀ = 1. For any M, let x' = 2πk, x'' = π + 2πk (k large).

|\cos(2\pi k) - \cos(\pi + 2\pi k)| = |1 - (-1)| = 2 > 1

Cauchy fails, so limit DNE. ∎

Example 3.15: Proving with Cauchy

Show $\lim_{x \to +\infty} 1/x$ exists.

For x', x'' > M = 2/ε:

\left|\frac{1}{x'} - \frac{1}{x''}\right| \leq \frac{1}{x'} + \frac{1}{x''} < \varepsilon

Remark 3.8: When to Use Cauchy

Prove existence without knowing the limit value
Prove non-existence for oscillating functions

7. Additional Properties

Theorem 3.13: Composite Limits

If $\lim_{x \to x_0} g(x) = L$ and f continuous at L, then $\lim f(g(x)) = f(L)$ .

Example 3.16: Composite

Find $\lim_{x \to 0} e^{\sin x}$ .

Since sin x → 0 and eˣ is continuous: $e^{\sin x} \to e^0 = 1$ .

Theorem 3.14: Absolute Value

If $\lim |f(x)| = 0$ , then $\lim f(x) = 0$ .

Example 3.17: Using |f|

$\lim_{x \to 0} x\cos(1/x) = 0$

$|x\cos(1/x)| \leq |x| \to 0$ , so limit = 0.

Example 3.18: Combined Techniques

Find $\lim_{x \to 0} \frac{\sin 3x}{\sin 2x}$ .

\frac{\sin 3x}{\sin 2x} = \frac{\sin 3x}{3x} \cdot \frac{2x}{\sin 2x} \cdot \frac{3}{2} \to 1 \cdot 1 \cdot \frac{3}{2} = \frac{3}{2}

Example 3.19: More Practice

Find $\lim_{x \to 0} \frac{1 - \cos x}{x \sin x}$ .

Use equivalences: 1 - cos x ~ x²/2, sin x ~ x

\frac{1 - \cos x}{x \sin x} \sim \frac{x^2/2}{x \cdot x} = \frac{1}{2}

8. More Worked Examples

Example 3.20: Polynomial Division

Find $\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$ .

Solution: Factor using difference of cubes:

x^3 - 1 = (x - 1)(x^2 + x + 1)

\lim_{x \to 1} \frac{(x-1)(x^2 + x + 1)}{x - 1} = \lim_{x \to 1} (x^2 + x + 1) = 1 + 1 + 1 = 3

Example 3.21: Higher Order Roots

Find $\lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x - 8}$ .

Solution: Let u = ∛x, so x = u³ and x → 8 means u → 2.

\frac{\sqrt[3]{x} - 2}{x - 8} = \frac{u - 2}{u^3 - 8} = \frac{u - 2}{(u-2)(u^2 + 2u + 4)} = \frac{1}{u^2 + 2u + 4}

As u → 2:

\lim = \frac{1}{4 + 4 + 4} = \frac{1}{12}

Example 3.22: Double Rationalization

Find $\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{x}$ .

Solution: Multiply by conjugate:

\frac{\sqrt{1+x} - \sqrt{1-x}}{x} \cdot \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}

= \frac{(1+x) - (1-x)}{x(\sqrt{1+x} + \sqrt{1-x})} = \frac{2x}{x(\sqrt{1+x} + \sqrt{1-x})} = \frac{2}{\sqrt{1+x} + \sqrt{1-x}}

As x → 0: $\frac{2}{1 + 1} = 1$

Example 3.23: Exponential and Trig

Find $\lim_{x \to 0} \frac{e^x - 1}{\sin x}$ .

Solution: Use standard limits:

\frac{e^x - 1}{\sin x} = \frac{e^x - 1}{x} \cdot \frac{x}{\sin x}

As x → 0: both factors → 1, so limit = 1.

Example 3.24: Logarithmic

Find $\lim_{x \to 1} \frac{\ln x}{x - 1}$ .

Solution: Let t = x - 1, so x = 1 + t and x → 1 means t → 0:

\frac{\ln x}{x - 1} = \frac{\ln(1 + t)}{t} \to 1 \text{ as } t \to 0

Example 3.25: Nested Functions

Find $\lim_{x \to 0} \frac{\tan(\sin x)}{x}$ .

\frac{\tan(\sin x)}{x} = \frac{\tan(\sin x)}{\sin x} \cdot \frac{\sin x}{x}

As x → 0: sin x → 0, so tan(sin x)/(sin x) → 1. Also sin x/x → 1.

Limit = 1 × 1 = 1.

Example 3.26: Power Form 1^∞

Find $\lim_{x \to 0} (1 + 2x)^{1/x}$ .

Solution: This is 1^∞ form. Rewrite:

(1 + 2x)^{1/x} = [(1 + 2x)^{1/(2x)}]^2

As x → 0, 2x → 0, so (1 + 2x)^{1/(2x)} → e.

Limit = e² .

Example 3.27: Form 0/0 with Trig

Find $\lim_{x \to 0} \frac{\tan x - \sin x}{x^3}$ .

\tan x - \sin x = \frac{\sin x}{\cos x} - \sin x = \sin x \cdot \frac{1 - \cos x}{\cos x}

Use sin x ~ x, 1 - cos x ~ x²/2, cos x → 1:

\frac{\sin x(1 - \cos x)}{x^3 \cos x} \sim \frac{x \cdot x^2/2}{x^3 \cdot 1} = \frac{1}{2}

Example 3.28: At Infinity

Find $\lim_{x \to +\infty} \left(\frac{x+1}{x-1}\right)^x$ .

Solution: This is 1^∞ form. Write:

\frac{x+1}{x-1} = 1 + \frac{2}{x-1}

\left(1 + \frac{2}{x-1}\right)^x = \left[\left(1 + \frac{2}{x-1}\right)^{(x-1)/2}\right]^{2x/(x-1)}

As x → +∞: (x-1)/2 → ∞, so inner part → e.

And 2x/(x-1) → 2. So limit = e².

Example 3.29: Squeeze with Absolute Value

Find $\lim_{x \to 0} x^2 \sin(1/x^2) \cos(1/x)$ .

Solution: Both sin and cos are bounded by [-1, 1]:

|x^2 \sin(1/x^2) \cos(1/x)| \leq x^2 \cdot 1 \cdot 1 = x^2 \to 0

By squeeze: limit = 0.

9. Detailed Proofs of Limit Laws

Proof of Product Rule: lim(fg) = (lim f)(lim g):

Let lim f = a and lim g = b. We show lim fg = ab.

Write:

f(x)g(x) - ab = f(x)g(x) - f(x)b + f(x)b - ab = f(x)[g(x)-b] + b[f(x)-a]

By local boundedness, |f(x)| ≤ M near x₀.

Given ε > 0, choose δ such that:

|f(x) - a| < ε/(2|b| + 1)
|g(x) - b| < ε/(2M)

|f(x)g(x) - ab| \leq M \cdot \frac{\varepsilon}{2M} + |b| \cdot \frac{\varepsilon}{2|b|+1} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon

∎

Proof of Quotient Rule: lim(f/g) = (lim f)/(lim g) when lim g ≠ 0:

First prove: if lim g = b ≠ 0, then lim(1/g) = 1/b.

By sign preservation, |g(x)| > |b|/2 near x₀.

\left|\frac{1}{g(x)} - \frac{1}{b}\right| = \frac{|b - g(x)|}{|b||g(x)|} < \frac{|g(x) - b|}{|b| \cdot |b|/2} = \frac{2|g(x) - b|}{b^2}

Given ε, choose δ: |g(x) - b| < εb²/2. Then |1/g - 1/b| < ε.

Then lim(f/g) = lim(f · 1/g) = (lim f)(lim 1/g) = a · (1/b) = a/b. ∎

∎

Remark 3.9: Why These Proofs Matter

Understand the logic behind computational rules
See how local boundedness and sign preservation are used
Be able to prove similar results on exams

10. Common Mistakes to Avoid

Mistake 1: Using Quotient Rule When Denominator → 0

Wrong: lim_{x→1} (x²-1)/(x-1) = (1-1)/(1-1) = 0/0 = ???

Correct: Factor first: (x²-1)/(x-1) = x+1 → 2.

Mistake 2: Assuming Strict Inequality Preserved

Wrong: Since x² > 0 for x ≠ 0, lim x² > 0.

Correct: We can only say lim x² ≥ 0. In fact, lim_{x→0} x² = 0.

Mistake 3: Incorrect Squeeze Bounds

Wrong: -1 ≤ sin(1/x) ≤ 1, so -1 ≤ x sin(1/x) ≤ 1.

Correct: Multiply by |x|: -|x| ≤ x sin(1/x) ≤ |x|.

Mistake 4: Confusing Local and Global Properties

Wrong: lim_{x→1} f(x) = 2 implies f is bounded on ℝ.

Correct: f is only guaranteed bounded NEAR x = 1, not everywhere.

Mistake 5: Forgetting Conditions

Wrong: lim f(g(x)) = f(lim g(x)) always.

Correct: Only when f is continuous at lim g(x).

11. Preview: Application to Continuity

The limit properties we've studied are essential for understanding continuity (Chapter 3.3) and its consequences.

Remark 3.10: How Limit Properties Connect to Continuity

Sum/Product: Sums and products of continuous functions are continuous
Quotient: Quotients are continuous where denominator ≠ 0
Composite: Compositions of continuous functions are continuous
Squeeze: Used to prove sin x/x → 1 (key for derivatives)

Example 3.30: Continuity Application

Show that f(x) = x² sin(1/x) for x ≠ 0, f(0) = 0 is continuous at 0.

Solution: By squeeze, lim_{x→0} x² sin(1/x) = 0.

Since f(0) = 0 = lim_{x→0} f(x), f is continuous at 0. ∎

12. Study Guide and Key Formulas

Quick Reference: Key Theorems

Uniqueness

If limit exists, it's unique

Local Boundedness

Limit exists ⟹ |f| ≤ M near x₀

Sign Preservation

L > 0 ⟹ f(x) > 0 near x₀

Order Preservation

f ≤ g ⟹ lim f ≤ lim g

Limit Laws Summary

\lim(f \pm g) = \lim f \pm \lim g

\lim(f \cdot g) = (\lim f)(\lim g)

\lim\frac{f}{g} = \frac{\lim f}{\lim g} \quad (\lim g \neq 0)

\lim f^n = (\lim f)^n

Problem-Solving Strategies

0/0 form: Factor, rationalize, or use L'Hôpital
∞/∞ form: Divide by highest power
Oscillating: Use squeeze theorem
1^∞ form: Rewrite using e^{ln}
Standard limits: sin x/x → 1, etc.

13. Extended Squeeze Theorem Applications

Example 3.31: Floor Function Limit

Find $\lim_{x \to +\infty} \frac{\lfloor x \rfloor}{x}$ .

Solution: Recall x - 1 < ⌊x⌋ ≤ x. Dividing by x (x > 0):

\frac{x-1}{x} < \frac{\lfloor x \rfloor}{x} \leq 1

1 - \frac{1}{x} < \frac{\lfloor x \rfloor}{x} \leq 1

As x → +∞: 1 - 1/x → 1. By Squeeze: limit = 1.

Example 3.32: Trigonometric Product

Find $\lim_{x \to 0} \frac{\sin x \sin 2x \sin 3x}{x^3}$ .

Solution: Write as product of standard limits:

\frac{\sin x \sin 2x \sin 3x}{x^3} = \frac{\sin x}{x} \cdot \frac{\sin 2x}{2x} \cdot 2 \cdot \frac{\sin 3x}{3x} \cdot 3

As x → 0: each sin(kx)/(kx) → 1.

\lim = 1 \cdot 1 \cdot 2 \cdot 1 \cdot 3 = 6

Example 3.33: Exponential Squeeze

Show $\lim_{x \to +\infty} \frac{x^n}{e^x} = 0$ for any n ∈ ℕ.

Proof: For x > 2n, we have eˣ > x^{2n}/(2n)! (from Taylor series).

0 < \frac{x^n}{e^x} < \frac{x^n (2n)!}{x^{2n}} = \frac{(2n)!}{x^n} \to 0

By Squeeze: limit = 0. Exponentials dominate polynomials. ∎

Example 3.34: Oscillation at Zero

Find $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) \cos\left(\frac{1}{x^2}\right)$ .

Solution: |sin(1/x)| ≤ 1 and |cos(1/x²)| ≤ 1, so:

|x^2 \sin(1/x) \cos(1/x^2)| \leq x^2 \to 0

By squeeze: limit = 0.

14. More Detailed Limit Law Examples

Example 3.35: Combining Multiple Laws

Find $\lim_{x \to 1} \frac{x^4 - 1}{x^3 - 1}$ .

Solution: Factor both:

x^4 - 1 = (x-1)(x+1)(x^2+1)

x^3 - 1 = (x-1)(x^2+x+1)

\frac{x^4-1}{x^3-1} = \frac{(x+1)(x^2+1)}{x^2+x+1}

At x = 1: $\frac{2 \cdot 2}{3} = \frac{4}{3}$

Example 3.36: Rational with Higher Powers

Find $\lim_{x \to 2} \frac{x^5 - 32}{x - 2}$ .

Method 1: Factor using a⁵ - b⁵ = (a-b)(a⁴ + a³b + a²b² + ab³ + b⁴):

x^5 - 32 = (x-2)(x^4 + 2x^3 + 4x^2 + 8x + 16)

At x = 2: 16 + 16 + 16 + 16 + 16 = 80.

Method 2: This is f'(2) where f(x) = x⁵ and derivative is 5x⁴ = 5(16) = 80.

Example 3.37: Double Limit

Find $\lim_{x \to 0} \frac{\sqrt{1 + \sin x} - 1}{x}$ .

Solution: Rationalize:

\frac{\sqrt{1+\sin x} - 1}{x} \cdot \frac{\sqrt{1+\sin x} + 1}{\sqrt{1+\sin x} + 1} = \frac{\sin x}{x(\sqrt{1+\sin x} + 1)}

= \frac{\sin x}{x} \cdot \frac{1}{\sqrt{1+\sin x} + 1} \to 1 \cdot \frac{1}{2} = \frac{1}{2}

Example 3.38: Trigonometric Identity

Find $\lim_{x \to 0} \frac{1 - \cos^3 x}{x \sin 2x}$ .

Solution: Factor 1 - cos³x = (1 - cos x)(1 + cos x + cos²x):

\frac{(1-\cos x)(1 + \cos x + \cos^2 x)}{x \cdot 2\sin x \cos x}

Use 1 - cos x ~ x²/2, sin x ~ x, cos x → 1:

\frac{(x^2/2)(3)}{x \cdot 2x \cdot 1} = \frac{3x^2/2}{2x^2} = \frac{3}{4}

15. Advanced Cauchy Criterion Applications

Example 3.39: Non-existence via Cauchy

Prove $\lim_{x \to +\infty} \sin(x^2)$ does not exist.

Proof: Take ε₀ = 1. For any M > 0, choose k such that √(2πk) > M.

Let x' = √(2πk) and x'' = √(2πk + π/2).

Then (x')² = 2πk and (x'')² = 2πk + π/2.

|\sin(2\pi k) - \sin(2\pi k + \pi/2)| = |0 - 1| = 1 \geq \varepsilon_0

Cauchy criterion fails, so limit DNE. ∎

Example 3.40: Existence via Cauchy

Prove $\lim_{x \to +\infty} \frac{\sin x}{x}$ exists using Cauchy.

Proof: For x', x'' > M:

\left|\frac{\sin x'}{x'} - \frac{\sin x''}{x''}\right| \leq \frac{1}{x'} + \frac{1}{x''} < \frac{2}{M}

Choose M = 2/ε. Then difference < ε. Cauchy is satisfied. ∎

Remark 3.11: Cauchy vs Direct Proofs

Cauchy criterion is useful when:

You don't know the limit value
Direct ε-δ seems difficult
You want to prove non-existence

For existence, Cauchy doesn't tell you WHAT the limit is—just that it exists.

16. How Properties Relate to Each Other

Property Dependency Map

ε-δ Definition → Uniqueness

Uniqueness is derived directly from ε-δ by contradiction

ε-δ Definition → Local Boundedness

Take ε = 1 in the definition

ε-δ Definition → Sign Preservation

Take ε = |L|/2 when L ≠ 0

Local Boundedness → Product Rule

Bounded factor controls the product

Sign Preservation → Quotient Rule

Ensures denominator bounded away from 0

Order Preservation → Squeeze Theorem

Squeeze is a special case of order preservation

17. Detailed Solution Techniques

Technique 1: Factoring

Key Formulas:

a² - b² = (a-b)(a+b)

a³ - b³ = (a-b)(a² + ab + b²)

a³ + b³ = (a+b)(a² - ab + b²)

aⁿ - bⁿ = (a-b)(aⁿ⁻¹ + aⁿ⁻²b + ... + bⁿ⁻¹)

Technique 2: Rationalization

Multiply by conjugate when you see √ in numerator or denominator:

\frac{\sqrt{a} - \sqrt{b}}{c} \cdot \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} + \sqrt{b}} = \frac{a - b}{c(\sqrt{a} + \sqrt{b})}

This eliminates radicals by creating a difference of squares.

Technique 3: Standard Limit Substitution

Replace with standard forms when approaching indeterminate:

\frac{\sin u}{u} \to 1 \text{ as } u \to 0

\frac{e^u - 1}{u} \to 1 \text{ as } u \to 0

\frac{\ln(1+u)}{u} \to 1 \text{ as } u \to 0

(1+u)^{1/u} \to e \text{ as } u \to 0

Technique 4: Equivalent Infinitesimals

As x → 0, these are equivalent (use in products only!):

sin x ~ x

tan x ~ x

arcsin x ~ x

arctan x ~ x

1 - cos x ~ x²/2

eˣ - 1 ~ x

ln(1+x) ~ x

(1+x)ᵅ - 1 ~ αx

Technique 5: Handling 1^∞ Form

When you see (1 + something)^{big}, use:

(1 + f(x))^{g(x)} = e^{g(x) \ln(1 + f(x))}

Then use ln(1 + u) ~ u when u → 0:

g(x) \ln(1 + f(x)) \approx g(x) \cdot f(x)

The limit becomes e^{lim g(x)f(x)}.

18. More Worked Examples

Example 3.41: Power Form

Find $\lim_{x \to 0} \left(\frac{\sin x}{x}\right)^{1/x^2}$ .

Solution: As x → 0, sin x/x → 1, so this is 1^∞ form.

Write as exponential:

\left(\frac{\sin x}{x}\right)^{1/x^2} = e^{\frac{1}{x^2} \ln\left(\frac{\sin x}{x}\right)}

Need to find: $\lim_{x \to 0} \frac{\ln(\sin x/x)}{x^2}$

Using Taylor: sin x/x = 1 - x²/6 + O(x⁴), so ln(sin x/x) ~ -x²/6.

\frac{\ln(\sin x/x)}{x^2} \to -\frac{1}{6}

Final answer: $e^{-1/6} = \frac{1}{\sqrt[6]{e}}$

Example 3.42: L'Hôpital Preview

Find $\lim_{x \to 0} \frac{x - \arctan x}{x^3}$ .

Solution: Taylor: arctan x = x - x³/3 + O(x⁵)

x - \arctan x = x - (x - \frac{x^3}{3} + O(x^5)) = \frac{x^3}{3} + O(x^5)

\frac{x - \arctan x}{x^3} = \frac{x^3/3 + O(x^5)}{x^3} = \frac{1}{3} + O(x^2) \to \frac{1}{3}

Example 3.43: Product of Limits

Find $\lim_{x \to 0} \frac{(e^x - 1)(\ln(1+x))}{x^2}$ .

\frac{(e^x - 1)(\ln(1+x))}{x^2} = \frac{e^x - 1}{x} \cdot \frac{\ln(1+x)}{x}

As x → 0: (eˣ - 1)/x → 1 and ln(1+x)/x → 1.

Limit = 1 · 1 = 1.

Example 3.44: At Negative Infinity

Find $\lim_{x \to -\infty} \sqrt{x^2 + x} + x$ .

Solution: For x < 0, √(x²) = |x| = -x.

\sqrt{x^2 + x} = |x|\sqrt{1 + 1/x} = -x\sqrt{1 + 1/x}

\sqrt{x^2 + x} + x = -x\sqrt{1 + 1/x} + x = x(1 - \sqrt{1 + 1/x})

Rationalize: multiply by (1 + √(1+1/x))/(1 + √(1+1/x)):

= \frac{x(1 - (1 + 1/x))}{1 + \sqrt{1 + 1/x}} = \frac{-1}{1 + \sqrt{1 + 1/x}} \to \frac{-1}{2}

Example 3.45: Mixed Forms

Find $\lim_{x \to 0^+} x^{x^x}$ .

Solution: First find lim xˣ as x → 0⁺.

x^x = e^{x \ln x} \to e^0 = 1 \text{ since } x\ln x \to 0

So x^{xˣ} → x¹ = x → 0 as x → 0⁺.

But wait—more carefully: x^{xˣ} = e^{xˣ ln x}.

As x → 0⁺: xˣ → 1 and ln x → -∞, but the product: xˣ · ln x → 1 · (-∞) = -∞.

So x^{xˣ} → e^{-∞} = 0.

19. Practice Problems

Problem Set A: Basic Limits

$\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$ (Answer: 6)
$\lim_{x \to 0} \frac{\sin 5x}{3x}$ (Answer: 5/3)
$\lim_{x \to 0} \frac{e^{3x} - 1}{2x}$ (Answer: 3/2)
$\lim_{x \to 0} \frac{\ln(1+4x)}{x}$ (Answer: 4)
$\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}$ (Answer: 1/2)

Problem Set B: Intermediate

$\lim_{x \to 0} \frac{1 - \cos 2x}{x \tan x}$ (Answer: 2)
$\lim_{x \to 0} (1 + 3x)^{2/x}$ (Answer: e⁶)
$\lim_{x \to +\infty} \left(\frac{2x+1}{2x-1}\right)^x$ (Answer: e)
$\lim_{x \to 0} \frac{\arctan x - x}{x^3}$ (Answer: -1/3)
$\lim_{x \to 0} x^2 \cot^2 x$ (Answer: 1)

Problem Set C: Advanced

$\lim_{x \to 0} \frac{\tan x - \arctan x}{x^3}$ (Answer: 2/3)
$\lim_{x \to 0^+} x^{\sin x}$ (Answer: 1)
$\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x}$ (Answer: 0)
$\lim_{x \to +\infty} (\sqrt{x^2+x} - x)$ (Answer: 1/2)
$\lim_{x \to 0} \frac{\sqrt[3]{1+x} - \sqrt[3]{1-x}}{x}$ (Answer: 2/3)

Problem Set D: Proof Problems

Prove uniqueness using ε-δ:
If lim f(x) = L₁ and lim f(x) = L₂, prove L₁ = L₂.
Prove local boundedness:
If lim f(x) = L, prove |f(x)| ≤ |L| + 1 near x₀.
Prove the product rule:
If lim f = a and lim g = b, prove lim(fg) = ab.
Use Cauchy criterion:
Prove lim_{x→∞} sin(x²) does not exist.

20. Key Insights and Tips

Insight 1: The Power of Substitution

When evaluating limits, clever substitutions can simplify problems dramatically:

Let u = 1/x to convert x → ∞ to u → 0
Let t = x - a to center the limit at 0
Let u = √x to handle square roots
Let u = eˣ to handle exponentials

Insight 2: Recognizing Standard Forms

Many limits reduce to standard forms in disguise:

$\frac{\sin 3x}{5x} = \frac{3}{5} \cdot \frac{\sin 3x}{3x}$ → (3/5)·1 = 3/5

$(1 + 2x)^{3/x} = [(1 + 2x)^{1/(2x)}]^6$ → e⁶

$\frac{e^{5x} - 1}{3x} = \frac{5}{3} \cdot \frac{e^{5x} - 1}{5x}$ → 5/3

Insight 3: When Things Don't Work

If direct methods fail, consider:

Squeeze theorem for oscillating bounded functions
Cauchy criterion to prove non-existence
Taylor series for complicated expressions
L'Hôpital's rule for 0/0 or ∞/∞ (Chapter 4)

Insight 4: Connection to Derivatives

Many of these limit techniques appear in derivatives:

f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}

Understanding limits is essential for understanding calculus!

21. Reference Tables

Indeterminate Forms

Form	Strategy
0/0	Factor, rationalize, L'Hôpital, Taylor
∞/∞	Divide by highest power, L'Hôpital
0 · ∞	Rewrite as 0/0 or ∞/∞
∞ - ∞	Combine fractions, factor, rationalize
1^∞	Use e^{g·ln(1+f)} with ln(1+u) ~ u
0⁰	Use e^{g·ln f}
∞⁰	Use e^{g·ln f}

Property Reference

Property	Statement	Use
Uniqueness	If limit exists, it's unique	Foundation
Local Boundedness	Limit ⟹ bounded nearby	Product proofs
Sign Preservation	L > 0 ⟹ f > 0 nearby	Quotient proofs
Order Preservation	f ≤ g ⟹ lim f ≤ lim g	Squeeze theorem
Squeeze	f ≤ g ≤ h, lim f = lim h = L ⟹ lim g = L	Oscillating limits
Cauchy	\|f(x') - f(x'')\| < ε near x₀	Existence proofs

Essential Standard Limits

Trigonometric

$\lim_{x \to 0} \frac{\sin x}{x} = 1$

$\lim_{x \to 0} \frac{\tan x}{x} = 1$

$\lim_{x \to 0} \frac{1-\cos x}{x^2} = \frac{1}{2}$

$\lim_{x \to 0} \frac{\arcsin x}{x} = 1$

$\lim_{x \to 0} \frac{\arctan x}{x} = 1$

Exponential & Logarithmic

$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$

$\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$

$\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$

$\lim_{x \to 0} (1+x)^{1/x} = e$

$\lim_{x \to \infty} (1+\frac{1}{x})^x = e$

Study Tips for Mastery

For Exam Preparation

• Memorize all standard limits
• Practice recognizing indeterminate forms
• Know when to use each technique
• Be able to write formal ε-δ proofs

Common Exam Questions

• Prove uniqueness of limits
• Prove product/quotient rules
• Use Cauchy to prove non-existence
• Evaluate limits using properties

What's Next?

In CALC-3.3, we'll use these limit properties to study:

Continuity at a point: lim f(x) = f(x₀)
Types of discontinuities: removable, jump, essential
Operations preserving continuity
Continuity of elementary functions

Chapter 3.2 Summary

Fundamental Properties

• Uniqueness: Limits are unique
• Local Boundedness: Limit ⟹ bounded near point
• Sign Preservation: L > 0 ⟹ f(x) > 0 nearby
• Order Preservation: f ≤ g ⟹ lim f ≤ lim g

Computational Tools

• Limit Laws: +, −, ×, ÷
• Squeeze Theorem: f ≤ g ≤ h with same limits
• Cauchy Criterion: Values cluster together
• Composite: f(g(x)) → f(lim g)

Practice Quiz: Limit Properties

Questions

Correct

Accuracy

\lim_{x \to x_0} f(x) = L

, what can we say?

Easy

Not attempted

\lim_{x \to x_0} f(x) = L > 0

, then near

x_0

Medium

Not attempted

What is

\lim_{x \to 0} x \sin(1/x)

Medium

Not attempted

Cauchy criterion: limit exists iff:

Hard

Not attempted

If lim f = a, lim g = b ≠ 0, then lim(f/g) =

Easy

Not attempted

Squeeze: if f ≤ g ≤ h and lim f = lim h = L, then:

Easy

Not attempted

FAQs

Local vs global boundedness?

Local: bounded near x₀. Global: bounded on entire domain. Limit implies local only.

When can't we use quotient rule?

When denominator limit is 0. Use L'Hôpital or other techniques then.

Strict inequality preserved?

No! f < g only implies lim f ≤ lim g, not strict inequality.