1.4 L'Hôpital's Rule, Curve Sketching, and Optimization

1.4 L'Hôpital's Rule, Curve Sketching, and Optimization

Differential calculus provides concrete machinery for resolving indeterminate limits, graphing curves through asymptotic analysis, and optimizing constrained objective functions. Mastery of these applications ensures rapid scoring on the GRE Mathematics Subject Test.

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Rigorous Application of L'Hôpital's Rule

Let $f, g$ be differentiable on an open interval $I$ containing $a$ (except possibly at $a$) with $g'(x) \neq 0$. If $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$, or $\lim_{x \to a} |g(x)| = \infty$, and $\lim_{x \to a} \frac{f'(x)}{g'(x)} = L \in [-\infty, \infty]$, then: $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} = L.$

Conversion of Indeterminate Forms

• Product $0 \cdot \infty$: Write $f \cdot g = \frac{f}{1/g}$ or $\frac{g}{1/f}$ to convert to $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
• Difference $\infty - \infty$: Combine over common denominators, rationalize with conjugates, or factor dominant terms.
• Exponential $1^\infty, 0^0, \infty^0$: Set $y = f(x)^{g(x)}$, compute $\ln y = g(x) \ln f(x)$, resolve the $0 \cdot \infty$ limit to find $L = \lim \ln y$, and conclude the original limit is $e^L$.

Two Critical L'Hôpital Pitfalls

1. Determinate Forms: Differentiating a non-indeterminate quotient yields errors. For example, $\lim_{x \to 0} \frac{\cos x}{x}$ is determinate ($1/0 \to \pm\infty$). Differentiating gives $\frac{-\sin x}{1} \to 0$, which is false.
2. Oscillatory Ratios: L'Hôpital is a one-way implication. For $\lim_{x \to \infty} \frac{x + \sin x}{x} = 1$, the derivative quotient $\frac{1 + \cos x}{1}$ oscillates without limit. L'Hôpital is inconclusive, yet the limit exists.

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Systematic Curve Sketching and Asymptotes

A complete curve sketch combines five structural elements:

1. Intercepts and Symmetry

• Symmetry: Even if $f(-x) = f(x)$ ($y$-axis reflection); Odd if $f(-x) = -f(x)$ (origin rotation); Periodic if $f(x + T) = f(x)$.
• Intercepts: $y$-intercept $(0, f(0))$ and roots $f(x) = 0$.

2. Complete Asymptote Taxonomy

• Vertical: Line $x = c$ where $\lim_{x \to c^\pm} f(x) = \pm\infty$.
• Horizontal: Line $y = L$ where $\lim_{x \to \infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$. Remember $\sqrt{x^2} = -x$ for $x < 0$, which often yields distinct horizontal asymptotes as $x \to \pm\infty$.
• Slant (Oblique): Line $y = mx + b$ ($m \neq 0$) where $\lim_{x \to \infty} [f(x) - (mx + b)] = 0$. For rational functions, slant asymptotes occur when $\deg(\text{num}) = \deg(\text{denom}) + 1$, found via polynomial long division.

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Applied Optimization and Related Rates

Optimization Workflow

1. Formulate objective function and constraint equations.
2. Eliminate variables to obtain single-variable function $Q = f(x)$ on feasible domain $I$.
3. Locate critical points via $f'(x) = 0$ or non-existence.
4. Classification: On $[a, b]$, compare critical values with boundary endpoints. On open or unbounded intervals, a critical point is global only after derivative-sign analysis or limiting behavior shows that all feasible values lie on the appropriate side; a single critical point alone is not enough.

Related Rates Workflow

1. Write the geometric equation linking variables $x(t), y(t), \theta(t)$.
2. Differentiate implicitly with respect to time $t$ via the chain rule.
3. Substitute numerical values only after taking all derivatives.

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Linear Approximations and Differentials

The linearization of $f$ at $a$ is $L(x) = f(a) + f'(a)(x - a)$, with differential $dy = f'(x) dx$.

• Taylor Remainder Bound: The error $R_1(x) = f(x) - L(x)$ satisfies: $|f(x) - L(x)| \le \frac{M}{2} |x - a|^2,$ where $M = \max |f''(t)|$ on the interval between $a$ and $x$.

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Step-by-Step Worked Problems

Problem 1: Indeterminate Exponential Limit

Evaluate $\lim_{x \to 0^+} (1 + 3x)^{2/x}$.

Solution:

1. Indeterminate form $1^\infty$. Set $y = (1 + 3x)^{2/x}$.
2. Take logarithms: $\ln y = \frac{2 \ln(1 + 3x)}{x}$, which is $\frac{0}{0}$ as $x \to 0^+$.
3. Apply L'Hôpital's Rule: $\lim_{x \to 0^+} \frac{2 \cdot \frac{3}{1 + 3x}}{1} = \lim_{x \to 0^+} \frac{6}{1 + 3x} = 6.$
4. Since $\lim \ln y = 6$, the original limit is $e^6$.

Problem 2: Inscribed Rectangle Optimization

Find the maximum area of a rectangle inscribed in the upper half of ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with base on the $x$-axis.

Solution:

1. Vertex in quadrant 1 is $(x, y)$ with $y = \frac{b}{a}\sqrt{a^2 - x^2}$ for $x \in [0, a]$.
2. Area $A(x) = 2x y = \frac{2b}{a} x \sqrt{a^2 - x^2}$.
3. Maximize $S(x) = [A(x)]^2 = \frac{4b^2}{a^2}(a^2 x^2 - x^4)$.
4. Differentiate: $S'(x) = \frac{4b^2}{a^2}(2a^2 x - 4x^3) = \frac{8b^2}{a^2} x (a^2 - 2x^2)$.
5. Setting $S'(x) = 0$ gives $x = a/\sqrt{2}$. Then $y = b/\sqrt{2}$.
6. Maximum Area: $A = 2(a/\sqrt{2})(b/\sqrt{2}) = ab$.

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GRE Exam Traps & Pitfalls

Trap 1: Freezing Variables Before Differentiating in Related Rates Never substitute static values (e.g., $x = 5$) before computing derivatives. Plugging values in too early makes variable rates erroneously vanish.

Trap 2: Overlooking Negative Root Signs in Horizontal Asymptotes For $f(x) = \frac{2x}{\sqrt{x^2 + 1}}$, as $x \to -\infty$, $\sqrt{x^2} = -x$, so dividing through produces a limit of $-2$, giving $y = -2$ alongside $y = +2$ as $x \to +\infty$.

Trap 3: Cycling L'Hôpital When Taylor Expansion Is Immediate When evaluating $\lim_{x \to 0} \frac{x - \sin x}{x^3}$, repeated differentiation is error-prone. The Taylor expansion $\sin x = x - x^3/6 + O(x^5)$ immediately gives $\frac{x^3/6}{x^3} = 1/6$.