Mathematics / Calculus I Single-Variable Differential Calculus 100% Free Open Access
Chapter 7 • Theory & Derivations

Mean Value Theorems, Taylor Approximations & L'Hôpital's Rule

The deep theoretical core of differential calculus: Rolle's theorem and proof via EVT and Fermat's theorem, Lagrange's Mean Value Theorem, zero-derivative constant function theorem, Cauchy generalized MVT, rigorous proof of L'Hopital's rule for 0/0 and infinity/infinity, resolution of exponential indeterminate forms, and linear differentials.

§7.1 Rolle's Theorem & Lagrange's Mean Value Theorem (MVT)

1. Rolle's Theorem

$$\mathbf{\text{Theorem (Rolle): Let } f: [a, b] \to \mathbb{R} \text{ satisfy:}}$$ $$\mathbf{1. } f \text{ is continuous on the closed interval } [a, b],$$ $$\mathbf{2. } f \text{ is differentiable on the open interval } (a, b), \text{ and}$$ $$\mathbf{3. } f(a) = f(b).$$ $$\mathbf{\text{Then there exists at least one number } c \in (a, b) \text{ such that } f'(c) = 0.}$$

Rigorous Proof: By the Extreme Value Theorem, continuous $f$ on $[a, b]$ attains an absolute maximum $M$ and minimum $m$.

  • Case 1 (Constant Function): If $M = m$, then $f(x)$ is constant on $[a, b]$. Thus $f'(x) = 0$ for every $x \in (a, b)$, and any $c \in (a, b)$ satisfies the theorem.
  • Case 2 (Non-Constant Function): Since $f(a) = f(b)$, at least one of the extrema (say the maximum $M$) must be attained at an interior point $c \in (a, b)$. Since $f$ is differentiable at $c$ and attains a local maximum, Fermat's Interior Extremum Theorem forces $f'(c) = 0 \quad \blacksquare$

2. Lagrange's Mean Value Theorem (MVT)

$$\mathbf{\text{Theorem (MVT): If } f \in C[a, b] \text{ and } f \in D(a, b), \text{ then there exists at least one } c \in (a, b) \text{ such that:}}$$ $$\mathbf{f'(c) = \frac{f(b) - f(a)}{b - a} \quad \iff \quad f(b) - f(a) = f'(c)(b - a)}$$

Geometric Meaning: There exists an interior point $c$ where the instantaneous rate of change (tangent slope) is strictly parallel to the average rate of change (secant slope connecting endpoints $(a, f(a))$ and $(b, f(b))$).

Proof via Rolle's Theorem: Construct the auxiliary function measuring the vertical distance between the curve and the secant line:

$$h(x) = f(x) - f(a) - \left[\frac{f(b) - f(a)}{b - a}\right](x - a)$$

Notice $h(a) = 0$ and $h(b) = 0$. Since $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, $h(x)$ satisfies all hypotheses of Rolle's Theorem on $[a, b]$. Thus $\exists c \in (a, b)$ with $h'(c) = 0$:

$$h'(c) = f'(c) - \frac{f(b) - f(a)}{b - a} = 0 \implies f'(c) = \frac{f(b) - f(a)}{b - a} \quad \blacksquare$$

§7.2 Fundamental Corollaries of the MVT & Monotonicity Criteria

1. The Zero-Derivative Constant Function Theorem

$$\mathbf{\text{Corollary 1: If } f'(x) = 0 \text{ for all } x \in (a, b), \text{ then } f(x) \text{ is constant on } (a, b).}$$

Proof: Choose any two points $x_1 < x_2$ in $(a, b)$. By the MVT on $[x_1, x_2]$, $\exists c \in (x_1, x_2)$ such that $f(x_2) - f(x_1) = f'(c)(x_2 - x_1) = 0 \cdot (x_2 - x_1) = 0$. Thus $f(x_1) = f(x_2)$ for all pairs, proving $f(x) \equiv C \quad \blacksquare$

2. Functions Differing by a Constant

$$\mathbf{\text{Corollary 2: If } f'(x) = g'(x) \text{ for all } x \in (a, b), \text{ then } f(x) = g(x) + C \text{ for some constant } C \in \mathbb{R}.}$$

This is the fundamental justification for the $+ C$ constant of integration in antiderivatives.

3. Monotonicity Test Criteria

$$\begin{aligned} f'(x) > 0 \quad \forall x \in (a, b) &\implies f \text{ is strictly increasing on } [a, b] \ f'(x) < 0 \quad \forall x \in (a, b) &\implies f \text{ is strictly decreasing on } [a, b] \end{aligned}$$

§7.3 Cauchy's Generalized Mean Value Theorem & Rigorous L'Hôpital's Rule

1. Cauchy's Generalized Mean Value Theorem

$$\mathbf{\text{Theorem: If } f, g \in C[a, b] \text{ and } f, g \in D(a, b) \text{ with } g'(x) \ne 0 \text{ on } (a, b), \text{ then there exists } c \in (a, b) \text{ such that:}}$$ $$\mathbf{\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}}$$

Proof: By Rolle's Theorem on $g(x)$, $g(b) \ne g(a)$ (otherwise $g'(c) = 0$). Construct $H(x) = [f(b) - f(a)]g(x) - [g(b) - g(a)]f(x)$. Since $H(a) = H(b) = f(b)g(a) - g(b)f(a)$, Rolle's theorem yields $H'(c) = 0 \implies [f(b)-f(a)]g'(c) = [g(b)-g(a)]f'(c) \quad \blacksquare$

2. L'Hôpital's Rule for Indeterminate Forms 0/0 and $\infty/\infty$

$$\mathbf{\text{Theorem: Suppose } \lim_{x \to c} f(x) = 0 \text{ and } \lim_{x \to c} g(x) = 0 \text{ (or both } \pm\infty\text{)}. \text{ If } \lim_{x \to c} \frac{f'(x)}{g'(x)} = L \text{ exists (or is } \pm\infty\text{)}, \text{ then:}}$$ $$\mathbf{\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} = L}$$

Proof for $0/0$: Define $f(c) = g(c) = 0$ to make $f, g$ continuous at $c$. For any $x$ near $c$, by Cauchy's MVT on $[c, x]$ (or $[x, c]$), there exists $x_1$ strictly between $c$ and $x$ such that:

$$\frac{f(x)}{g(x)} = \frac{f(x) - f(c)}{g(x) - g(c)} = \frac{f'(x_1)}{g'(x_1)}$$

As $x \to c$, $x_1 \to c$ by squeezing. Thus $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x_1 \to c} \frac{f'(x_1)}{g'(x_1)} = L \quad \blacksquare$

3. Resolution of Other Indeterminate Forms

  • Product $0 \cdot \infty$: Rewrite $f \cdot g = \frac{f}{1/g}$ or $\frac{g}{1/f}$ to convert to $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
  • Difference $\infty - \infty$: Find a common algebraic denominator or rationalize.
  • Exponential Forms $0^0, 1^\infty, \infty^0$: Let $y = [f(x)]^{g(x)}$, take logarithms $\ln y = g(x) \ln f(x)$ (which becomes $0 \cdot \infty$), evaluate limit $L$, and recover $\lim y = e^L$.

§7.4 Linear Approximations, Differentials & Error Propagation

1. Linear (Tangent Line) Approximation

The tangent line to $y = f(x)$ at $x = x_0$ provides the optimal first-order linear approximation $L(x) \approx f(x)$ for values of $x$ close to $x_0$:

$$L(x) = f(x_0) + f'(x_0)(x - x_0)$$

2. Differentials

Let $y = f(x)$. If $dx = \Delta x$ represents an independent change in the input, the differential $dy$ represents the corresponding change along the tangent line:

$$dy = f'(x) dx$$

While the actual change along the curve is $\Delta y = f(x + \Delta x) - f(x)$, for small $dx$ we have $\Delta y \approx dy$.

3. Error Estimation Metrics

  • Absolute Error: $\Delta y \approx dy = f'(x_0) dx$
  • Relative Error: $\frac{\Delta y}{y} \approx \frac{dy}{y} = \frac{f'(x_0) dx}{f(x_0)}$
  • Percentage Error: $\left( \frac{dy}{y} \right) \times 100\%$
TIERED UNIVERSITY HONORS PROBLEMS

Step-by-Step Solved Examination Problems

Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.

Tier 1: Foundational Example 7.1: L'Hôpital Evaluation of Exponential Indeterminate Form 1^∞

Evaluate the exact limit using L'Hôpital's Rule: $\lim_{x \to 0} (1 + 3x)^{2/x}$.

Step 1: Identify Indeterminate Form and Take Logarithm
$$\text{As } x \to 0, \quad (1 + 3(0))^{2/0} = [1^\infty]. \quad \text{Let } y = (1 + 3x)^{2/x}$$

Taking natural log of both sides converts power to product:

Step 2: Transform into Fraction for L'Hôpital
$$\ln y = \frac{2}{x} \ln(1 + 3x) = \frac{2\ln(1 + 3x)}{x}$$

As $x o 0$, numerator is $2\ln(1) = 0$ and denominator is $0$, which is the indeterminate form $0/0$.

Step 3: Apply L'Hôpital's Rule
$$\lim_{x \to 0} \ln y = \lim_{x \to 0} \frac{\frac{d}{dx}[2\ln(1 + 3x)]}{\frac{d}{dx}[x]} = \lim_{x \to 0} \frac{2 \left(\frac{3}{1 + 3x}\right)}{1} = \frac{6}{1 + 0} = 6$$

Differentiating numerator and denominator independently.

Step 4: Exponentiate to Recover Limit of $y$
$$\lim_{x \to 0} y = e^{\lim_{x \to 0} \ln y} = e^6$$

Since the natural exponential function is continuous, $\lim e^{\ln y} = e^{\lim \ln y}$.

Final Answer & Physical Insight

\lim_{x \to 0} (1 + 3x)^{2/x} = e^6

Tier 2: Intermediate Exam Example 7.2: Application of Rolle's Theorem to Prove Uniqueness of Real Roots

Prove that the polynomial equation $f(x) = 2x^5 + 5x^3 + 10x - 4 = 0$ has exactly one real root in $\mathbb{R}$. (Use the IVT for existence, and Rolle's Theorem for uniqueness).

Step 1: Prove Existence via Intermediate Value Theorem
$$f(0) = -4 < 0, \qquad f(1) = 2(1) + 5(1) + 10(1) - 4 = 13 > 0$$

Since $f$ is continuous on $[0, 1]$ and $f(0) f(1) < 0$, the IVT guarantees at least one root $c \in (0, 1)$.

Step 2: Compute Derivative and Analyze Sign
$$f'(x) = 10x^4 + 15x^2 + 10 = 5(2x^4 + 3x^2 + 2)$$

Notice that for all real $x \in \mathbb{R}$, $x^4 \ge 0$ and $x^2 \ge 0$. Thus $f'(x) \ge 10 > 0$ strictly for every $x \in \mathbb{R}$.

Step 3: Uniqueness Proof by Contradiction via Rolle's Theorem
$$\text{Suppose } \exists x_1 < x_2 \text{ such that } f(x_1) = f(x_2) = 0$$

Since $f$ is a polynomial, it is continuous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$. By Rolle's Theorem, there must exist $c \in (x_1, x_2)$ such that $f'(c) = 0$.

Step 4: Conclude Contradiction
$$f'(c) = 10c^4 + 15c^2 + 10 \ge 10 \ne 0 \implies \text{Contradiction!}$$

Since $f'(x)$ is never zero, it is impossible for two distinct roots to exist. Therefore, exactly one real root exists in $\mathbb{R}$.

Final Answer & Physical Insight

f(x) = 0 \text{ has exactly one real root in } (0, 1) \subset \mathbb{R}

Tier 3: Honors / Proof Challenge Example 7.3: Cauchy MVT Proof of Classical Strict Inequality Bounds on sin x

Use the Mean Value Theorem and Cauchy's MVT to prove rigorously that for all $x > 0$: $\cos x > 1 - \frac{x^2}{2}$ and $\sin x > x - \frac{x^3}{6}$.

Step 1: First MVT Application to $\sin t$ on $[0, x]$
$$\frac{\sin x - \sin 0}{x - 0} = \cos(c_1) \implies \sin x = x \cos(c_1) < x \cdot 1 = x \quad (0 < c_1 < x)$$

Since $\cos(c_1) < 1$ for $c_1 \in (0, x)$, we establish $\sin x < x$ for all $x > 0$.

Step 2: Second MVT Application to Auxiliary Function for Cosine
$$g(t) = \cos t - \left(1 - \frac{t^2}{2}\right) \implies g'(t) = -\sin t + t = t - \sin t$$

From Step 1, $g'(t) = t - \sin t > 0$ for all $t > 0$.

Step 3: Deduce Cosine Inequality by Monotonicity
$$g(x) - g(0) = g'(c_2)(x - 0) > 0 \implies \cos x - \left(1 - \frac{x^2}{2}\right) > 0 \implies \cos x > 1 - \frac{x^2}{2}$$

Since $g(0) = 1 - 1 = 0$, $g(x) > 0$ strictly for all $x > 0$.

Step 4: Third Application to Establish Sine Lower Bound
$$h(t) = \sin t - \left(t - \frac{t^3}{6}\right) \implies h'(t) = \cos t - \left(1 - \frac{t^2}{2}\right) > 0$$

By Step 3, $h'(t) > 0$ for all $t > 0$. Since $h(0) = 0$, $h(x) > 0$ for all $x > 0$, proving $\sin x > x - \frac{x^3}{6} \quad \blacksquare$.

Final Answer & Physical Insight

\cos x > 1 - \frac{x^2}{2} \quad \text{and} \quad \sin x > x - \frac{x^3}{6} \quad \forall x > 0 \quad (\text{Strict Taylor Bounds})