4.1 Multivariable Functions, Limits, and Partial Derivatives

4.1 Multivariable Functions, Limits, and Partial Derivatives

Multivariable differential calculus extends real analysis to scalar fields $f: D \subset \mathbb{R}^n \to \mathbb{R}$. On the GRE Mathematics Subject Test, questions evaluate both computational proficiency—such as partial derivatives and tangent planes—and analytical subtleties, including path-dependent limits, mixed partial symmetry, and the gap between partial derivatives and total differentiability.

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Multivariable Functions and Level Curves

For a two-variable function $z = f(x, y)$, the geometric graph forms a surface in $\mathbb{R}^3$. Spatial geometry is captured in the plane by level curves: $L_k(f) = \{(x, y) \in D : f(x, y) = k\}$ Plotting level curves at uniform intervals $\Delta k$ produces a contour map. Closely spaced contours indicate steep gradients; widely spaced contours indicate relatively flat terrain. For $w = F(x, y, z)$, constant levels $F(x, y, z) = k$ define level surfaces in $\mathbb{R}^3$.

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Multivariable Limits and the Two-Path Test

The limit $\lim_{(x, y) \to (x_0, y_0)} f(x, y) = L$ means: $\forall \varepsilon > 0, \; \exists \delta > 0 \text{ such that } 0 < \|\mathbf{x} - \mathbf{x}_0\| < \delta \implies |f(\mathbf{x}) - L| < \varepsilon$

The Two-Path Test for Non-Existence

In $\mathbb{R}^2$, approach occurs along infinitely many continuous trajectories. If $f(x, y)$ approaches distinct limits along two different paths, the limit does not exist.

• Coordinate Axes: Test $x = 0$ ($y \to 0$) and $y = 0$ ($x \to 0$).
• Linear Rays: Test $y = mx$. If the limit depends on $m$, the limit does not exist.
• Parabolic Trajectories: When powers are unbalanced (e.g., $x^4 + y^2$), lines $y = mx$ give $0$, but $y = kx^2$ balances degrees and reveals path-dependence.

Proving Existence via Polar Coordinates

Path testing cannot prove existence. To prove a limit at $(0, 0)$, substitute $x = r \cos \theta, y = r \sin \theta$. If $|f(r \cos \theta, r \sin \theta) - L| \le g(r)$ where $\lim_{r \to 0^+} g(r) = 0$ independently of $\theta$, then the limit is $L$ by the Squeeze Theorem.

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Continuity in $\mathbb{R}^2$ and $\mathbb{R}^3$

A function $f$ is continuous at $(x_0, y_0)$ if $f(x_0, y_0)$ is defined, $\lim_{(x, y) \to (x_0, y_0)} f(x, y)$ exists, and their values coincide. Polynomials are continuous on $\mathbb{R}^n$; rational functions are continuous on their domains. By the Extreme Value Theorem (EVT), continuous functions on compact sets attain absolute extrema.

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Partial Derivatives and Clairaut's Theorem

First-order partial derivatives are single-variable derivatives holding remaining variables constant: $f_x(x_0, y_0) = \lim_{h \to 0} \frac{f(x_0 + h, y_0) - f(x_0, y_0)}{h}, \quad f_y(x_0, y_0) = \lim_{k \to 0} \frac{f(x_0, y_0 + k) - f(x_0, y_0)}{k}$ Geometrically, $f_x(x_0, y_0)$ is the slope of the tangent to the trace curve in the slicing plane $y = y_0$.

Clairaut's Theorem (Schwarz's Theorem)

If mixed second partials $f_{xy} = \frac{\partial^2 f}{\partial y \partial x}$ and $f_{yx} = \frac{\partial^2 f}{\partial x \partial y}$ are continuous on an open disk containing $(x_0, y_0)$, then: $f_{xy}(x_0, y_0) = f_{yx}(x_0, y_0)$ Equality fails if mixed partials are discontinuous (e.g., $f(x, y) = \frac{xy(x^2 - y^2)}{x^2 + y^2}$ with $f(0, 0) = 0$, where $f_{xy}(0, 0) = -1 \neq 1 = f_{yx}(0, 0)$).

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Differentiability vs. Partial Derivatives

In several variables, the existence of partial derivatives $f_x, f_y$ does not imply continuity or differentiability.

Total Differentiability

$f(x, y)$ is differentiable at $(x_0, y_0)$ if: $\lim_{(\Delta x, \Delta y) \to (0, 0)} \frac{f(x_0 + \Delta x, y_0 + \Delta y) - f(x_0, y_0) - f_x \Delta x - f_y \Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} = 0$

Regularity Comparison

Tangent Planes and Linearization

For a differentiable function $f$, the tangent plane at $(x_0, y_0, z_0)$ is: $z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$ The linearization is $L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)$.

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Step-by-Step Worked Problems

Problem 1: Two-Path Test on Degree-Unbalanced Limits

Evaluate $\lim_{(x, y) \to (0, 0)} \frac{x^2 y}{x^4 + y^2}$.

Solution:

1. Linear paths $y = mx$: $\lim_{x \to 0} \frac{x^2(mx)}{x^4 + m^2 x^2} = \lim_{x \to 0} \frac{mx}{x^2 + m^2} = 0$.
2. Parabolic paths $y = kx^2$: $\lim_{x \to 0} \frac{x^2(kx^2)}{x^4 + k^2 x^4} = \frac{k}{1 + k^2}$.
3. For $k = 1$, the limit is $1/2$; for $k = -1$, it is $-1/2$.
4. Different paths yield different limits, so the limit does not exist.

Problem 2: Tangent Plane Computation

Find the tangent plane to $f(x, y) = \sqrt{20 - x^2 - 7y^2}$ at $(2, 1)$.

Solution:

1. Point value: $z_0 = f(2, 1) = \sqrt{20 - 4 - 7} = 3$.
2. Partials: $f_x = \frac{-x}{\sqrt{20 - x^2 - 7y^2}} \implies f_x(2, 1) = -\frac{2}{3}$; $f_y = \frac{-7y}{\sqrt{20 - x^2 - 7y^2}} \implies f_y(2, 1) = -\frac{7}{3}$.
3. Plane: $z - 3 = -\frac{2}{3}(x - 2) - \frac{7}{3}(y - 1) \iff 2x + 7y + 3z = 20$.

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GRE Exam Traps & Pitfalls

Trap 1: Checking Only Straight Lines ($y = mx$) Showing $\lim_{x \to 0} f(x, mx) = L$ does not prove limit existence. Check parabolic paths $y = kx^2$ when denominator powers are unequal.

Trap 2: Assuming Partial Derivatives Imply Continuity The existence of $f_x$ and $f_y$ does not guarantee continuity at $(x_0, y_0)$.

Trap 3: Omitting the Continuity Hypothesis in Clairaut's Theorem The equality $f_{xy} = f_{yx}$ requires second mixed partials to be continuous.