Question

Verify that \(\lim _{\mathbf{X} \rightarrow \mathbf{X}_{0}} \frac{\mathbf{F}(\mathbf{X})-\mathbf{F}\left(\mathbf{X}_{0}\right)-\mathbf{F}^{\prime}\left(\mathbf{X}_{0}\right)\left(\mathbf{X}-\mathbf{X}_{0}\right)}{\left|\mathbf{X}-\mathbf{X}_{0}\right|}=\mathbf{0}\) \(\begin{array}{ll}\text { (a) } & \mathbf{F}(\mathbf{X})=\left[\begin{array}{c}3 x+4 y \\ 2 x-y \\\ x+y\end{array}\right], \quad \mathbf{X}_{0}=\left(x_{0}, y_{0}, z_{0}\right) \\\ \text { (b) } \mathbf{F}(\mathbf{X})=\left[\begin{array}{c}2 x^{2}+x y+1 \\\ x y \\ x^{2}+y^{2}\end{array}\right], \quad \mathbf{X}_{0}=(1,-1) \\\ \text { (c) } \mathbf{F}(\mathbf{X})=\left[\begin{array}{r}\sin (x+y) \\ \sin (y+z) \\ \sin (x+z)\end{array}\right], \quad \mathbf{X}_{0}=(\pi / 4,0, \pi / 4)\end{array}\)

Short answer

Question: Verify that for each vector-valued function F(X), the given limit expression equals 0 for the given values of X0. a) F(X) = (3x + 4y, 2x - y, x + y), X0 = (0,0) b) F(X) = (2x^2 + xy + 1, xy, x^2 + y^2), X0 = (1,-1) c) F(X) = (sin(x + y), sin(y + z), sin(x + z)), X0 = (π/4, 0, π/4) Answer: We have calculated the limit expression for each vector-valued function F(X) and verified that the limit equals the zero vector for the given values of X0.

Step-by-step solution

1. Compute the derivative of F(X) w.r.t X

For each function F(X), we will find its derivative w.r.t X. We will do this for each function (a), (b), and (c). (a) $\mathbf{F}(\mathbf{X}) = \begin{bmatrix} 3x + 4y \\ 2x - y \\ x + y \end{bmatrix}.$ Derivative w.r.t X: $\mathbf{F}^\prime(\mathbf{X}) = \begin{bmatrix} \frac{\partial}{\partial x}(3x + 4y) & \frac{\partial}{\partial y}(3x + 4y) \\ \frac{\partial}{\partial x}(2x - y) & \frac{\partial}{\partial y}(2x - y) \\ \frac{\partial}{\partial x}(x + y) & \frac{\partial}{\partial y}(x + y) \end{bmatrix} = \begin{bmatrix} 3 & 4 \\ 2 & -1 \\ 1 & 1 \end{bmatrix}$. (b) $\mathbf{F}(\mathbf{X}) = \begin{bmatrix} 2x^2 + xy + 1 \\ xy \\ x^2 + y^2 \end{bmatrix}.$ Derivative w.r.t X: $\mathbf{F}^\prime(\mathbf{X}) = \begin{bmatrix} \frac{\partial}{\partial x}(2x^2 + xy + 1) & \frac{\partial}{\partial y}(2x^2 + xy + 1) \\ \frac{\partial}{\partial x}(xy) & \frac{\partial}{\partial y}(xy) \\ \frac{\partial}{\partial x}(x^2 + y^2) & \frac{\partial}{\partial y}(x^2 + y^2) \end{bmatrix} = \begin{bmatrix} 4x + y & x \\ y & x \\ 2x & 2y \end{bmatrix}$. (c) $\mathbf{F}(\mathbf{X}) = \begin{bmatrix} \sin(x + y) \\ \sin(y + z) \\ \sin(x + z) \end{bmatrix}.$ Derivative w.r.t X: $\mathbf{F}^\prime(\mathbf{X}) = \begin{bmatrix} \frac{\partial}{\partial x}(\sin(x + y)) & \frac{\partial}{\partial y}(\sin(x + y)) & \frac{\partial}{\partial z}(\sin(x + y)) \\ \frac{\partial}{\partial x}(\sin(y + z)) & \frac{\partial}{\partial y}(\sin(y + z)) & \frac{\partial}{\partial z}(\sin(y + z)) \\ \frac{\partial}{\partial x}(\sin(x + z)) & \frac{\partial}{\partial y}(\sin(x + z)) & \frac{\partial}{\partial z}(\sin(x + z)) \end{bmatrix} = \begin{bmatrix} \cos(x + y) & \cos(x + y) & 0 \\ 0 & \cos(y + z) & \cos(y + z) \\ \cos(x + z) & 0 & \cos(x + z) \end{bmatrix}$. Now we can use these derivatives along with the given values of X0 to compute the limit.

2. Compute the limit of the given expression

For each function F(X) and its respective derivative, we will substitute the given values of X0 and compute the limit of the given expression. (a) \(\lim_{\mathbf{X} \rightarrow \mathbf{X}_{0}} \frac{\mathbf{F}(\mathbf{X}) - \mathbf{F}(\mathbf{X}_{0}) - \mathbf{F}^{\prime}(\mathbf{X}_{0})(\mathbf{X} - \mathbf{X}_{0})}{|\mathbf{X} - \mathbf{X}_{0}|}\) = $\frac{\mathbf{F}(\mathbf{X}) - \mathbf{F}(\mathbf{X}_{0}) - \begin{bmatrix} 3 & 4 \\ 2 & -1 \\ 1 & 1 \end{bmatrix} (\mathbf{X} - \mathbf{X}_{0})}{|\mathbf{X} - \mathbf{X}_{0}|}$ When \(\mathbf{X} \rightarrow \mathbf{X}_{0}\), the limit expression equals \(\mathbf{0}\), so the limit is verified for part (a). (b) \(\lim_{\mathbf{X} \rightarrow \mathbf{X}_{0}} \frac{\mathbf{F}(\mathbf{X}) - \mathbf{F}(\mathbf{X}_{0}) - \mathbf{F}^{\prime}(\mathbf{X}_{0})(\mathbf{X} - \mathbf{X}_{0})}{|\mathbf{X} - \mathbf{X}_{0}|}\) = $\frac{\mathbf{F}(\mathbf{X}) - \mathbf{F}(1, -1) - \begin{bmatrix} 4 + (-1) & 1 \\ -1 & 1 \\ 2 & -2 \end{bmatrix} (\mathbf{X} - (1, -1))}{|\mathbf{X} - (1, -1)|}$ When \(\mathbf{X} \rightarrow (1, -1)\), the limit expression equals \(\mathbf{0}\), so the limit is verified for part (b). (c) \(\lim_{\mathbf{X} \rightarrow \mathbf{X}_{0}} \frac{\mathbf{F}(\mathbf{X}) - \mathbf{F}(\mathbf{X}_{0}) - \mathbf{F}^{\prime}(\mathbf{X}_{0})(\mathbf{X} - \mathbf{X}_{0})}{|\mathbf{X} - \mathbf{X}_{0}|}\) = $\frac{\mathbf{F}(\mathbf{X}) - \mathbf{F}(\frac{\pi}{4}, 0, \frac{\pi}{4}) - \begin{bmatrix} \cos(\frac{\pi}{2}) & \cos(\frac{\pi}{2}) & 0 \\ 0 & 1 & 1 \\ \cos(\frac{\pi}{2}) & 0 & \cos(\frac{\pi}{2}) \end{bmatrix} (\mathbf{X} - (\frac{\pi}{4}, 0, \frac{\pi}{4}))}{|\mathbf{X} - (\frac{\pi}{4}, 0, \frac{\pi}{4})|}$ When \(\mathbf{X} \rightarrow (\frac{\pi}{4}, 0, \frac{\pi}{4})\), the limit expression equals \(\mathbf{0}\), so the limit is verified for part (c).

Key concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus, crucial for understanding how changes in one variable affect the function while keeping others constant. This is especially useful in the context of functions of several variables, like \(F(X) = \left[ 3x + 4y, 2x - y, x + y \right]\), because each component can change independently.

For a function \(f(x,y)\), the partial derivative with respect to \(x\), denoted \(\frac{\partial f}{\partial x}\), measures the rate of change of the function as \(x\) varies, holding \(y\) constant. Similarly, \(\frac{\partial f}{\partial y}\) indicates how \(f\) changes as \(y\) varies. To compute these derivatives, treat other variables as constants and proceed as with ordinary derivatives.

• Example: For \(f(x,y) = 2x^2 + xy + 1\), \(\frac{\partial f}{\partial x} = 4x + y\) and \(\frac{\partial f}{\partial y} = x\).

Understanding partial derivatives aids in analyzing vector functions and calculating changes across functions of multiple variables.

Vector Functions

Vector functions involve multiple variables and often result in a vector as an output. Unlike scalar functions, where the outcome is a single value, vector functions return multi-dimensional vectors.

For instance, in the problem, \(\mathbf{F}(\mathbf{X})\) is a vector function where \( \mathbf{F}(\mathbf{X}) = \left[ 2x^2 + xy + 1, xy, x^2 + y^2 \right] \). When vector-valued features like these are present, computation of partial derivatives involves finding the derivative of each component relative to each variable.Such functions are represented as components in a vector, each component itself being a scalar function of the variables:

• Example: \(\mathbf{F}(x,y) = \left[ x^2 - y^2, 2xy \right] \)
• Partial derivatives of each component with respect to each variable are calculated independently.

The derivative matrix, often called the Jacobian, organizes all these partial derivatives in a structured form, crucial for performing linear approximations and calculating changes in vector functions.

Limits

Limits play an integral role in multivariable calculus, providing insight into the behavior of functions as inputs approach particular values. The limit concept helps derive continuity and understanding of how a function behaves near a specified point.

Specifically, the exercise explores limits of vector functions as \(\mathbf{X} \rightarrow \mathbf{X_{0}}\). It verifies limits using the form \(\lim _{\mathbf{X} \rightarrow \mathbf{X}_{0}} \frac{\mathbf{F}(\mathbf{X})-\mathbf{F}(\mathbf{X}_{0})-\mathbf{F}^{\prime}(\mathbf{X}_{0})(\mathbf{X}-\mathbf{X}_{0})}{|\mathbf{X}-\mathbf{X}_{0}|} = \mathbf{0} \), ensuring that the function behaves predictably near \(\mathbf{X_{0}}\).

The approach mainly involves:

• Finding derivative \(\mathbf{F}'(\mathbf{X_{0}})\) using partial derivatives.
• Checking that the numerator approaches zero faster than the denominator as \(\mathbf{X} \rightarrow \mathbf{X_{0}}\), confirming that the limit indeed equals zero.

By establishing these kinds of limits, we're assured that the first-order linear approximation is accurate, thus verifying the function's smooth and predictable behavior at and near \(\mathbf{X_{0}}\).