Let `a= lim _(x rarr 1) (x/(lnx)-1/(xln x)), b = lim _(x rarr 0) ((x^(3)-16x)/(4x+x^(2))),`
`c= lim _(x rarr 1) ((ln(1+sinx))/x) & `
`d = lim _(x rarr -1) ((x+1)^(3))/([sin (x+1) - (x+1)])`
Then `[[a,b],[c,d]]` is

Options

A. idempotent
B. involutory
C. non-singular
D. nilpotent

Explanation

To solve the given problem, we need to evaluate the limits \( a \), \( b \), \( c \), and \( d \) step by step.

### Step 1: Calculate \( a = \lim_{x \to 1} \left( \frac{x}{\ln x} - \frac{1}{x \ln x} \right) \)

1. **Combine the fractions**:
\[ a = \lim_{x \to 1} \left( \frac{x^2 - 1}{x \ln x} \right) \]
Here, we take the common denominator \( x \ln x \).

2. **Evaluate the limit**:
When \( x \to 1 \), both the numerator and denominator approach 0, resulting in the indeterminate form \( \frac{0}{0} \). We apply L'Hôpital's Rule:
\[ a = \lim_{x \to 1} \frac{2x}{\ln x + 1} \]

3. **Substituting \( x = 1 \)**:
\[ a = \frac{2 \cdot 1}{\ln 1 + 1} = \frac{2}{0 + 1} = 2 \]

### Step 2: Calculate \( b = \lim_{x \to 0} \frac{x^3 - 16x}{4x + x^2} \)

1. **Simplify the expression**:
\[ b = \lim_{x \to 0} \frac{x(x^2 - 16)}{x(4 + x)} = \lim_{x \to 0} \frac{x^2 - 16}{4 + x} \]

2. **Evaluate the limit**:
Again, substituting \( x = 0 \) gives \( \frac{-16}{4} = -4 \).

### Step 3: Calculate \( c = \lim_{x \to 1} \frac{\ln(1 + \sin x)}{x} \)

1. **Substituting \( x = 1 \)**:
\[ c = \frac{\ln(1 + \sin 1)}{1} = \ln(1 + \sin 1) \]

### Step 4: Calculate \( d = \lim_{x \to -1} \frac{(x + 1)^3}{\sin(x + 1) - (x + 1)} \)

1. **Evaluate the limit**:
This limit is also of the form \( \frac{0}{0} \). We apply L'Hôpital's Rule:
\[ d = \lim_{x \to -1} \frac{3(x + 1)^2}{\cos(x + 1) - 1} \]

2. **Substituting \( x = -1 \)** gives \( \frac{0}{0} \) again, so we differentiate again:
\[ d = \lim_{x \to -1} \frac{6(x + 1)}{-\sin(x + 1)} = \lim_{x \to -1} \frac{6(0)}{-\sin(0)} = \frac{6}{0} = -6 \]

### Final Matrix

Now we can substitute the values of \( a \), \( b \), \( c \), and \( d \) into the matrix:

\[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 2 & -4 \\ \ln(1 + \sin 1) & -6 \end{bmatrix} \]