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
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)
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:
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 $h'(c) = 0$:
§7.2 Fundamental Corollaries of the MVT & Monotonicity Criteria
1. The Zero-Derivative Constant Function Theorem
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
This is the fundamental justification for the $+ C$ constant of integration in antiderivatives.
3. Monotonicity Test Criteria
§7.3 Cauchy's Generalized Mean Value Theorem & Rigorous L'Hôpital's Rule
1. Cauchy's Generalized Mean Value Theorem
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 $\infty/\infty$
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:
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 $x_0$:
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:
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: $\left( \frac{dy}{y} \right) \times 100\%$
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.
Evaluate the exact limit using L'Hôpital's Rule: $\lim_{x \to 0} (1 + 3x)^{2/x}$.
Taking natural log of both sides converts power to product:
As $x o 0$, numerator is $2\ln(1) = 0$ and denominator is $0$, which is the indeterminate form $0/0$.
Differentiating numerator and denominator independently.
Since the natural exponential function is continuous, $\lim e^{\ln y} = e^{\lim \ln y}$.
\lim_{x \to 0} (1 + 3x)^{2/x} = e^6
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).
Since $f$ is continuous on $[0, 1]$ and $f(0) f(1) < 0$, the IVT guarantees at least one root $c \in (0, 1)$.
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 $x \in \mathbb{R}$.
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 $f'(c) = 0$.
Since $f'(x)$ is never zero, it is impossible for two distinct roots to exist. Therefore, exactly one real root exists in $\mathbb{R}$.
f(x) = 0 \text{ has exactly one real root in } (0, 1) \subset \mathbb{R}
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 $\sin x > x - \frac{x^3}{6}$.
Since $\cos(c_1) < 1$ for $c_1 \in (0, x)$, we establish $\sin x < x$ for all $x > 0$.
From Step 1, $g'(t) = t - \sin t > 0$ for all $t > 0$.
Since $g(0) = 1 - 1 = 0$, $g(x) > 0$ strictly for all $x > 0$.
By Step 3, $h'(t) > 0$ for all $t > 0$. Since $h(0) = 0$, $h(x) > 0$ for all $x > 0$, proving $\sin x > x - \frac{x^3}{6} \quad \blacksquare$.
\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})