Lecture 7: Intermediate Value Theorem, Squeeze Theorem, and Trigonometric Limits

The Intermediate Value Theorem (IVT)

• Rationale and Origin: The Intermediate Value Theorem was first proposed by Bolzano of Bohemia.

• Nature of the Theorem: The IVT is a property shared by all continuous functions.

• Primary Application: It is generally used in computations to find the values of x-intercepts where a function $y = f(x)$ intersects the x-axis. These values are where $f(x) = 0$ and are also known as the zeros or solutions of the equation $f(x) = 0$.

• Example of Zeros: If $c_1, c_2, c_3$ are zeros of a third-degree polynomial, then:

  • $f(x) = (x - c_1)(x - c_2)(x - c_3)$

  • $f(c_1) = 0, f(c_2) = 0, f(c_3) = 0$

  • Specifically, if $f(x) = (x - 4)(x - 1)(x - 3)$, the zeros are $4, 1, 3$.

Formal Statement of the Intermediate Value Theorem

• Hypothesis:

  • Suppose $f$ is a continuous function of the variable $x$ on a closed interval $[a, b]$ on the real number line $\mathbb{R}$.

  • Let the sign of $f(a)$ be different from the sign of $f(b)$, such that $f(a) \neq 0$ and $f(b) \neq 0$.

  • In addition, suppose $f(a) \neq f(b)$.

• Definition of N: Let $N$ be a real number between $f(a)$ and $f(b)$ such that:

  • f(a) < N < f(b) OR f(b) < N < f(a).

• Conclusion: There exists at least one $x$-value, say $x = c$, in the open interval $(a, b)$ (where a < c < b) such that $f(c) = N$.

Illustrations and Cases of the IVT

• Case #1: Increasing Scenario

  • Condition: f(a) < N and f(b) > N.

  • Because $f(x)$ is continuous on $[a, b]$, there exists at least one $x = c$ between $x = a$ and $x = b$ such that $f(c) = N$.

• Case #2: Decreasing Scenario

  • Condition: f(a) > N and f(b) < N.

  • Because $f(x)$ is continuous on $[a, b]$, there exists at least one $x = c$ between $x = a$ and $x = b$ such that $f(c) = N$.

• Case #3: Multiple Values of c

  • Condition: f(a) < N, f(b) > N, and $f(x)$ is continuous on $[a, b]$.

  • In some cases, the function may oscillate, resulting in multiple points $c_1, c_2, c_3, \dots$ in the interval $(a, b)$ where $f(c_1) = f(c_2) = f(c_3) = N$.

• Case #4: Root Finding ($N = 0$)

  • This specific case is used to solve polynomial equations of the form $f(x) = 0$.

  • It serves to estimate the roots/zeros of a function, especially useful when zeros are irrational numbers.

  • Reformulation for Zeros:

    • Let $y = f(x)$ be continuous on $[a, b]$.

    • Suppose there exists values $x = c_1$ and $x = c_2$ where f(c_1) > 0 and f(c_2) < 0, or vice versa.

    • Then there exists an $x$-value $x = c_0$ such that a \leq c_1 < c_0 < c_2 \leq b where $f(c_0) = 0$.

Strategy for Locating Zeros

• Calculator Utilization: It is possible to use features on the TI-83 PLUS and TI-84 PLUS to locate zeros.

• Methods:

  • TABLE feature: Explore the table to find where the sign of $f(x)$ changes.

  • Intercept feature: Directly calculate the zeros using graphing capabilities.

Worked Examples on the IVT

• Example 1: Consider $f(x) = 4x^3 - 6x^2 + 3x - 2$.

  • (i) Show there is a root between $x = 1$ and $x = 2$:

    • Step 1: $f(x)$ is a polynomial of degree $3$. All polynomials are continuous on $\mathbb{R}$, thus it is continuous on $[1, 2]$.

    • Step 2: Using the TABLE feature of a TI-84/83 PLUS calculator:

      • At $x = 1$, $f(1) = 4(1)^3 - 6(1)^2 + 3(1) - 2 = -1$.

      • At $x = 2$, $f(2) = 4(2)^3 - 6(2)^2 + 3(2) - 2 = 12$.

    • Step 3: Since f(1) = -1 < 0 and f(2) = 12 > 0, and the function is continuous, by the IVT there exists a value $c \in (1, 2)$ such that $f(c) = 0$.

  • (ii) Locate other zeros: By inspecting the graph on the calculator, it is found that there is only one value of $x = c$ where $f(x) = 0$.

• Example 2: Consider $f(x) = x^5 - 3x^3 + x^2 - 23x + 19$.

  • Step 1: Construct a table of values (via calculator or manual calculation):

    • $x = -4, f(x) = -705$

    • $x = -3, f(x) = -65$

    • $x = -2, f(x) = 61$

    • $x = 0, f(x) = 19$

    • $x = 1, f(x) = -5$

    • $x = 2, f(x) = -15$

    • $x = 3, f(x) = 121$

    • $x = 4, f(x) = 775$

  • Step 2: Identify sign changes across given intervals:

    • (i) Interval $[-3, -2]$: $f(-3) = -65$ and $f(-2) = 61$. Sign change exists; there is a zero $c_1 \in (-3, -2)$.

    • (ii) Interval $[0, 1]$: $f(0) = 19$ and $f(1) = -5$. Sign change exists; there is a zero $c_2 \in (0, 1)$.

    • (iii) Interval $[2, 3]$: $f(2) = -15$ and $f(3) = 121$. Sign change exists; there is a zero $c_3 \in (2, 3)$.

• Example 3: Consider $f(x) = x^3 - 4x + 1$.

  • Step 1: Construct Table:

    • $x = -4, f(x) = -47$

    • $x = -3, f(x) = -14$

    • $x = -2, f(x) = 1$

    • $x = -1, f(x) = 4$

    • $x = 0, f(x) = 1$

    • $x = 1, f(x) = -2$

    • $x = 2, f(x) = 1$

    • $x = 3, f(x) = 16$

  • Step 2: The function is a polynomial, thus continuous on $\mathbb{R}$. Check intervals:

    • (i) $[-3, -2]$: $f(-3) = -14$, $f(-2) = 1$. Sign change occurs; root $x = c_1$ exists in $(-3, -2)$.

    • (ii) $[0, 1]$: $f(0) = 1$, $f(1) = -2$. Sign change occurs; root $x = c_2$ exists in $(0, 1)$.

    • (iii) $[1, 2]$: $f(1) = -2$, $f(2) = 1$. Sign change occurs; root $x = c_3$ exists in $(1, 2)$.

The Squeeze Theorem

• Alternative Names: Also known as the Sandwich Theorem, the Pinch Theorem, or the Hamburger Theorem.

• Definition: Let $f, g,$ and $h$ be real-valued functions of the variable $x$ such that:

  1. $x = x_0$ is in the domains of $f(x), g(x),$ and $h(x)$, though the functions may or may not be defined exactly at $x = x_0$.

  2. In the neighborhood of $x_0$, $g(x) \leq f(x) \leq h(x)$.

  3. $\lim_{x \to x_0} g(x) = L$ and $\lim_{x \to x_0} h(x) = L$.

• Conclusion: $\lim_{x \to x_0} f(x) = L$.

Worked Examples on the Squeeze Theorem

• Example 1: Compute $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right)$.

  • Observation: $\sin\left(\frac{1}{x}\right)$ oscillates infinitely between $-1$ and $1$ as $x \to 0$. $\lim_{x \to 0} \sin\left(\frac{1}{x}\right)$ does not exist.

  • Step 1: Use the property $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$.

  • Step 2: Multiply by $|x|$ to get $-|x| \leq x \sin\left(\frac{1}{x}\right) \leq |x|$.

  • Step 3: Let $g(x) = -|x|$ and $h(x) = |x|$.

  • Step 4: Compute limits of bounds: $\lim_{x \to 0} -|x| = 0$ and $\lim_{x \to 0} |x| = 0$.

  • Conclusion: By the Squeeze Theorem, $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0$.

• Example 2: Compute $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$.

  • Step 1: Identify bounds: $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$.

  • Step 2: Multiply by $x^2$: $-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$.

  • Step 3: Let $g(x) = -x^2$ and $h(x) = x^2$.

  • Step 4: Since $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} x^2 = 0$, then by the Squeeze Theorem, $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$.

Trigonometric Limits

• Angular variables: Variables like $x, \theta, t$ are measured in radians and serve as dummy variables.

• Fundamental Limit:     $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$

• Sequence Evaluation for $\sin(x)/x$:

  • At $x = \pm 0.20$, $f(x) \approx 0.99334665$

  • At $x = \pm 0.10$, $f(x) \approx 0.99833417$

  • At $x = \pm 0.05$, $f(x) \approx 0.99958339$

  • At $x = \pm 0.01$, $f(x) \approx 0.99998333$

  • At $x = \pm 0.005$, $f(x) \approx 0.99999583$

  • At $x = \pm 0.001$, $f(x) \approx 0.99999983$

  • Note: While the function is undefined (DNE) at $x = 0$, it is irrelevant for the limit calculation.

• Inference: By inspection of the table, as $x \to 0^+$ and $x \to 0^-$, the limit is $1$.

• Generalizations for $n \in \mathbb{N}$:

  • $\lim_{x \to 0} \frac{\sin(2x)}{2x} = 1$

  • $\lim_{x \to 0} \frac{\sin(3x)}{3x} = 1$

  • $\lim_{x \to 0} \frac{\sin(nx)}{nx} = 1$

  • $\lim_{x \to 0} \frac{\sin(nx)}{x} = n$

• Another Fundamental Limit:     $\lim_{x \to 0} \frac{\cos(x) - 1}{x} = 0$     Similarly: $\lim_{x \to 0} \frac{1 - \cos(x)}{x} = 0$.

Basic Trigonometric Limits Summary

1. $\lim_{x \to 0} \sin(x) = 0$; $\lim_{x \to 0} \sin(ax) = 0$

2. $\lim_{x \to 0} \cos(x) = 1$; $\lim_{x \to 0} \cos(ax) = 1$

3. $\lim_{x \to 0} \tan(x) = 0$ because $\lim_{x \to 0} \frac{\sin(x)}{\cos(x)} = \frac{0}{1} = 0$.

4. $\lim_{x \to 0} \frac{\sin(ax)}{ax} = 1$

5. $\lim_{x \to a} \frac{\sin(x - a)}{x - a} = 1$

6. $\lim_{x \to -a} \frac{\sin(x + a)}{x + a} = 1$

7. $\lim_{x \to \infty} \frac{\sin(x)}{x} = 0$

Examples: Evaluating Trigonometric Limits

• Ex 1: Find $\lim_{x \to 2} \frac{\sin(x - 2)}{x^2 - 4}$.

  • Tactical rewriting: $f(x) = \frac{\sin(x - 2)}{(x - 2)(x + 2)} = \left[ \frac{\sin(x - 2)}{x - 2} \right] \times \left[ \frac{1}{x + 2} \right]$.

  • Limit: $\lim_{x \to 2} \left[ \frac{\sin(x - 2)}{x - 2} \right] \times \lim_{x \to 2} \left[ \frac{1}{x + 2} \right] = (1) \times \left( \frac{1}{2 + 2} \right) = \frac{1}{4}$.

• Ex 2: Evaluate $\lim_{x \to 0} \frac{\sin(3x)}{x}$.

  • Rewrite as $3 \times \frac{\sin(3x)}{3x}$.

  • Limit: $3 \times \lim_{x \to 0} \frac{\sin(3x)}{3x} = 3 \times (1) = 3$.

• Ex 3: Evaluate $\lim_{x \to 0} \frac{\sin(2x)}{\sin(3x)}$.

  • Tactical rewrite: $f(x) = \frac{\frac{\sin(2x)}{x}}{\frac{\sin(3x)}{x}} = \frac{2 \left(\frac{\sin(2x)}{2x}\right)}{3 \left(\frac{\sin(3x)}{3x}\right)}$.

  • Limit: $\frac{2 \lim_{x \to 0} \frac{\sin(2x)}{2x}}{3 \lim_{x \to 0} \frac{\sin(3x)}{3x}} = \frac{2(1)}{3(1)} = \frac{2}{3}$.

• Ex 4: Evaluate $\lim_{x \to 0} \ln\left[ \frac{\sin(x)}{x} \right]$.

  • $\ln\left[ \lim_{x \to 0} \frac{\sin(x)}{x} \right] = \ln(1) = 0$.

• Ex 5: Evaluate $\lim_{x \to 0} \frac{\tan(x)}{x}$.

  • Rewrite: $\frac{\sin(x)}{\cos(x) \cdot x} = \left(\frac{\sin(x)}{x}\right) \times \left(\frac{1}{\cos(x)}\right)$.

  • Limit: $(1) \times \left(\frac{1}{\cos(0)}\right) = 1 \times 1 = 1$.

• Ex 6: Evaluate $\lim_{x \to \infty} \frac{\sin(1/x)}{\sin(1/x)}$? (Transcript text unclear, re-evaluating): Let $f(x) = \frac{\sin(2/x)}{\sin(1/x)}$.

  • Rewrite: $f(x) = \frac{\frac{\sin(2/x)}{1/x}}{\frac{\sin(1/x)}{1/x}} = \frac{2 \frac{\sin(2/x)}{2/x}}{\frac{\sin(1/x)}{1/x}}$.

  • As $x \to \infty$, $1/x \to 0$. Therefore, limit is $\frac{2(1)}{1} = 2$.

• Ex 7: Evaluate $\lim_{x \to 0} \frac{\tan(3x)}{\cos(3x)}$.

  • Substitute directly: $\frac{\tan(0)}{\cos(0)} = \frac{0}{1} = 0$.

• Ex 8: Find $\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta + \tan(\theta)}$.

  • Divide numerator and denominator by $\sin(\theta)$:         $f(\theta) = \frac{1}{\frac{\theta}{\sin(\theta)} + \frac{\tan(\theta)}{\sin(\theta)}} = \frac{1}{\frac{\theta}{\sin(\theta)} + \frac{1}{\cos(\theta)}}$.

  • Limit: $\frac{1}{1 + 1} = \frac{1}{2}$.

• Ex 9: Compute $\lim_{x \to 0} \frac{\tan(x)}{\sin(2x)}$.

  • Rewrite: $f(x) = \frac{\frac{\sin(x)}{\cos(x)}}{2 \sin(x) \cos(x)} = \frac{1}{2 \cos^2(x)}$.

  • Alternative rewrite: $f(x) = \frac{\tan(x)}{x} \times \frac{x}{\sin(2x)} = (1) \times \frac{1}{2 \left(\frac{\sin(2x)}{2x}\right)}$.

  • Limit: $1 \times \frac{1}{2(1)} = \frac{1}{2}$.

• Ex 10: Evaluate $\lim_{x \to 0} \frac{1 - \cos(x)}{2x^2}$.

  • Trigonometric identity: $1 - \cos(x) = 2 \sin^2(x/2)$.

  • $f(x) = \frac{2 \sin^2(x/2)}{2x^2} = \frac{\sin^2(x/2)}{x^2} = \frac{\sin^2(x/2)}{4(x/2)^2} = \frac{1}{4} \left( \frac{\sin(x/2)}{x/2} \right)^2$.

  • Limit: $\frac{1}{4} (1)^2 = \frac{1}{4}$.

Limits at Infinity

• Rational Functions: Let $f(x) = \frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials.

  • $P(x) = a_n x^n + \dots + a_1 x + a_0$

  • $Q(x) = b_m x^m + \dots + b_1 x + b_0$

  • Degree of $P(x) = n$, degree of $Q(x) = m$.

• Algorithm to find $\lim_{x \to \infty} f(x)$:

  • Step 1: Check degrees $n$ and $m$.

  • Step 2: Divide both the numerator and denominator by $x^{\max(n, m)}$ and simplify.

  • Step 3: Apply the rule that $\lim_{x \to \pm \infty} \frac{1}{x^n} = 0$ for n > 0.

Worked Examples: Limits at Infinity

• Example 1: $f(x) = \frac{3x - 1}{3x + 1}$. Find $\lim_{x \to \infty} f(x)$.

  • Divide by $x$: $\frac{3 - 1/x}{3 + 1/x}$.

  • Limit: $\frac{3 - 0}{3 + 0} = 1$.

• Example 2: $f(x) = \frac{e^x - 1}{e^x + 1}$. Find $\lim_{x \to \infty} f(x)$.

  • Divide by $e^x$: $\frac{1 - 1/e^x}{1 + 1/e^x}$.

  • As $x \to \infty$, $1/e^x \to 0$. Limit: $\frac{1 - 0}{1 + 0} = 1$.

• Example 3: $f(x) = \frac{x^2 + 2x + 1}{x^3 - 4x + 2}$. Find $\lim_{x \to \infty} f(x)$.

  • Divide by $x^3$: $\frac{1/x + 2/x^2 + 1/x^3}{1 - 4/x^2 + 2/x^3}$.

  • Limit: $\frac{0 + 0 + 0}{1 - 0 + 0} = \frac{0}{1} = 0$.

• Example 4: $f(x) = \frac{3x^3 - 5x^2 + 4x - 8}{x^3 - 4x + 2}$. Find $\lim_{x \to \infty} f(x)$.

  • Divide by $x^3$: $\frac{3 - 5/x + 4/x^2 - 8/x^3}{1 - 4/x^2 + 2/x^3}$.

  • Limit: $\frac{3 - 0 + 0 - 0}{1 - 0 + 0} = 3$.

• Example 5: $f(x) = \frac{5x^4 - 14x + 2}{4x^3 + 8x + 1}$. Compute $\lim_{x \to \infty} f(x)$.

  • Divide by $x^4$: $\frac{5 - 14/x^3 + 2/x^4}{4/x + 8/x^3 + 1/x^4}$.

  • Limit: $\frac{5 - 0 + 0}{0 + 0 + 0} = \frac{5}{0} = \infty$.

Limits Involving Radical Signs and Infinity

• Important Relationship: $\sqrt{x^2} = |x|$.

  • If $x \to \infty$, then $|x| = x$.

  • If $x \to -\infty$, then $|x| = -x$.

• Example: $f(x) = \frac{\sqrt{2x^2 + 4}}{x + 5}$.

  • (i) Limit as $x \to \infty$:

    • The answer will be positive because both numerator and denominator stay positive.

    • Squaring and putting under root: $\sqrt{\frac{2x^2 + 4}{(x + 5)^2}} = \sqrt{\frac{2x^2 + 4}{x^2 + 10x + 25}}$.

    • Divide interior rational function by $x^2$: $\sqrt{\frac{2 + 4/x^2}{1 + 10/x + 25/x^2}}$.

    • Limit: $\sqrt{\frac{2 + 0}{1 + 0 + 0}} = \sqrt{2}$.

  • (ii) Limit as $x \to -\infty$:

    • As $x \to -\infty$, the numerator $\sqrt{2x^2+4}$ is positive, but the denominator $x+5$ is negative. The final sign will be negative.

    • Logic: $\frac{\sqrt{2x^2+4}}{x+5} = \frac{\sqrt{x^2}\sqrt{2 + 4/x^2}}{x(1 + 5/x)} = \frac{|x|\sqrt{2 + 4/x^2}}{x(1 + 5/x)}$.

    • Since x < 0, $|x|/x = -1$.

    • Limit: $-1 \times \sqrt{2} = -\sqrt{2}$.

Limits of Sums/Differences Involving Radicals

• Technique: Rationalize the numerator using the identity $(a - b)(a + b) = a^2 - b^2$.

• Example: $f(x) = \sqrt{x^2 + 3x} - x$.

  • (i) Limit as $x \to \infty$:

    • Multiple the term by $\frac{\sqrt{x^2 + 3x} + x}{\sqrt{x^2 + 3x} + x}$.

    • $f(x) = \frac{(x^2 + 3x) - x^2}{\sqrt{x^2 + 3x} + x} = \frac{3x}{\sqrt{x^2 + 3x} + x}$.

    • Divide numerator and denominator by $x$: $\frac{3}{\frac{\sqrt{x^2 + 3x}}{x} + 1} = \frac{3}{\sqrt{\frac{x^2 + 3x}{x^2}} + 1} = \frac{3}{\sqrt{1 + 3/x} + 1}$.

    • Limit: $\frac{3}{\sqrt{1 + 0} + 1} = \frac{3}{2}$.

  • (ii) Limit as $x \to -\infty$:

    • Following the rationalization, we get $f(x) = \frac{3x}{\sqrt{x^2 + 3x} + x}$.

    • Divide by $x$. Remember for x < 0, $x = -\sqrt{x^2}$. Thus $\frac{\sqrt{x^2+3x}}{x} = -\sqrt{1 + 3/x}$.

    • Limit function: $\frac{3}{-\sqrt{1 + 3/x} + 1}$.

    • Limit: $\frac{3}{-1 + 1} = \frac{3}{0} = \infty$.

• Example: Standard Conditions

  • (i) What conditions ensure $\lim_{x \to \infty} (\sqrt{ax^2 + bx + c} - \sqrt{dx^2 + ex + f})$ exists?

    • Condition: $a = d$.

  • (ii) What is the limit?

    • Result: $\frac{b - e}{2\sqrt{a}}$.

The Signum Function

• Definition: The signum function, denoted $f(x) = \text{sgn}(x)$, is defined as:

  • $\text{sgn}(x) = 1$ if x > 0

  • $\text{sgn}(x) = 0$ if $x = 0$

  • $\text{sgn}(x) = -1$ if x < 0

• Composite Signum Function: For $F(x) = \text{sgn}(g(x))$:

  • $1$ if g(x) > 0

  • $0$ if $g(x) = 0$

  • $-1$ if g(x) < 0

• Worked Example: Let $f(x) = \text{sgn}\left(\frac{x - 1}{x + 2}\right)$.

  • Piecewise analysis: Use a schematic diagram for $x = 1$ and $x = -2$.

    • Interval $(-\infty, -2)$: $\frac{\text{neg}}{\text{neg}} = \text{pos}$. Thus $f(x) = 1$.

    • Interval $(-2, 1)$: $\frac{\text{neg}}{\text{pos}} = \text{neg}$. Thus $f(x) = -1$.

    • Interval $(1, \infty)$: $\frac{\text{pos}}{\text{pos}} = \text{pos}$. Thus $f(x) = 1$.

    • At $x = 1$: $f(1) = 0$.

  • Continuity:

    • Domain: $(-\infty, -2) \cup (-2, \infty)$.

    • The function is discontinuous at $x = 1$ because $\lim_{x \to 1^-} f(x) = -1$ and $\lim_{x \to 1^+} f(x) = 1$.

    • Note that $x = -2$ is not a point of discontinuity because it is not in the domain of the function.