Calculate lim x->0 (1/(sin x) - 1/x)

The given limit is $\lim_{x \to 0} \left(\frac{1}{\sin x} - \frac{1}{x}\right)$.
First, combine the fractions by finding a common denominator:

$$\frac{1}{\sin x} - \frac{1}{x} = \frac{x - \sin x}{x \sin x}$$

As $x \to 0$, the numerator $x - \sin x \to 0 - \sin(0) = 0$, and the denominator $x \sin x \to 0 \cdot \sin(0) = 0$. This is an indeterminate form $\frac{0}{0}$, so we apply L'Hôpital's Rule.

We differentiate the numerator and the denominator with respect to $x$:
Derivative of numerator: $\frac{d}{dx}(x - \sin x) = 1 - \cos x$.
Derivative of denominator: $\frac{d}{dx}(x \sin x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x$.
So the limit becomes:

$$\lim_{x \to 0} \frac{1 - \cos x}{\sin x + x \cos x}$$

As $x \to 0$, the new numerator $1 - \cos x \to 1 - \cos(0) = 1 - 1 = 0$, and the new denominator $\sin x + x \cos x \to \sin(0) + 0 \cdot \cos(0) = 0 + 0 = 0$. This is still an indeterminate form $\frac{0}{0}$, so we apply L'Hôpital's Rule again.

We differentiate the new numerator and denominator with respect to $x$:
Derivative of numerator: $\frac{d}{dx}(1 - \cos x) = \sin x$.
Derivative of denominator: $\frac{d}{dx}(\sin x + x \cos x) = \cos x + (1 \cdot \cos x + x \cdot (-\sin x)) = \cos x + \cos x - x \sin x = 2 \cos x - x \sin x$.
So the limit becomes:

$$\lim_{x \to 0} \frac{\sin x}{2 \cos x - x \sin x}$$

Now, substitute $x = 0$ into the expression:

$$\frac{\sin(0)}{2 \cos(0) - 0 \cdot \sin(0)} = \frac{0}{2 \cdot 1 - 0} = \frac{0}{2} = 0$$

The limit is 0.

The correct option is A.

$$\boxed{\text{A) } 0}$$

Given matrices $A = \begin{pmatrix} 1 & 3 & -1 \\ -2 & 5 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 5 \\ 2 \\ 4 \end{pmatrix}$.
To find $AB$, we multiply the rows of $A$ by the column of $B$:

$$AB = \begin{pmatrix} (1)(5) + (3)(2) + (-1)(4) \\ (-2)(5) + (5)(2) + (3)(4) \end{pmatrix}$$ $$AB = \begin{pmatrix} 5 + 6 - 4 \\ -10 + 10 + 12 \end{pmatrix}$$ $$AB = \begin{pmatrix} 7 \\ 12 \end{pmatrix}$$

Given $AB = \begin{pmatrix} 7 \\ 12 \end{pmatrix}$ and $C = (2 \ -3)$.
To find $(AB)C$, we multiply the $2 \times 1$ matrix $AB$ by the $1 \times 2$ matrix $C$. The result will be a $2 \times 2$ matrix:

$$(AB)C = \begin{pmatrix} 7 \\ 12 \end{pmatrix} (2 \ -3)$$ $$(AB)C = \begin{pmatrix} (7)(2) & (7)(-3) \\ (12)(2) & (12)(-3) \end{pmatrix}$$ $$(AB)C = \begin{pmatrix} 14 & -21 \\ 24 & -36 \end{pmatrix}$$

This matches option A.

$$\boxed{\text{A) } \begin{pmatrix} 14 & -21 \\ 24 & -36 \end{pmatrix}}$$

The equation of a plane passing through a point $P(x_0, y_0, z_0)$ with a normal vector $\mathbf{n} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ is given by:

$$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$$

Given point $P(2, -1, -1)$, so $x_0 = 2$, $y_0 = -1$, $z_0 = -1$.
Given normal vector $\mathbf{n} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$, so $a = 2$, $b = -3$, $c = 1$.
Substitute these values into the equation:

$$2(x - 2) + (-3)(y - (-1)) + 1(z - (-1)) = 0$$ $$2(x - 2) - 3(y + 1) + 1(z + 1) = 0$$

Distribute the coefficients:

$$2x - 4 - 3y - 3 + z + 1 = 0$$

Combine the constant terms:

$$2x - 3y + z + (-4 - 3 + 1) = 0$$ $$2x - 3y + z - 6 = 0$$

The equation of the plane is $2x - 3y + z - 6 = 0$.

$$\boxed{2x - 3y + z - 6 = 0}$$