Let `a =lim _(xto0) (ln (cos 2x ))/( 3x ^(2)) , b = lim _(xto0) (sin ^(2) 2x )/(x (1-e ^(x))), c = lim _(x to 1) (sqrtx-x )/( ln x).` Then a,b,c satisfy :

Options

A. `a lt b lt c`
B. `b lt c lt a`
C. `a lt c lt b`
D. `b lt a lt c`

Answer

To solve the given limits step by step, we will evaluate each limit \( a \), \( b \), and \( c \) one by one.

### Step 1: Evaluate \( a = \lim_{x \to 0} \frac{\ln(\cos(2x))}{3x^2} \)

1. **Substituting \( x = 0 \)**:
\[ \ln(\cos(0)) = \ln(1) = 0 \quad \text{and} \quad 3(0)^2 = 0 \]
This gives us the indeterminate form \( \frac{0}{0} \).

2. **Apply L'Hôpital's Rule**:
Differentiate the numerator and denominator:
\[ \text{Numerator: } \frac{d}{dx}[\ln(\cos(2x))] = \frac{-\sin(2x) \cdot 2}{\cos(2x)} = -2\tan(2x) \]
\[ \text{Denominator: } \frac{d}{dx}[3x^2] = 6x \]
Thus, we have:
\[ a = \lim_{x \to 0} \frac{-2\tan(2x)}{6x} = \lim_{x \to 0} \frac{-\tan(2x)}{3x} \]

3. **Using the limit \( \lim_{x \to 0} \frac{\tan(kx)}{kx} = 1 \)**:
\[ a = \frac{-2}{3} \cdot 1 = -\frac{2}{3} \]

### Step 2: Evaluate \( b = \lim_{x \to 0} \frac{\sin^2(2x)}{x(1 - e^x)} \)

1. **Substituting \( x = 0 \)**:
\[ \sin^2(0) = 0 \quad \text{and} \quad 1 - e^0 = 0 \]
This gives us the indeterminate form \( \frac{0}{0} \).

2. **Apply L'Hôpital's Rule**:
Differentiate the numerator and denominator:
\[ \text{Numerator: } \frac{d}{dx}[\sin^2(2x)] = 2\sin(2x)\cdot 2\cos(2x) = 4\sin(2x)\cos(2x) = 2\sin(4x) \]
\[ \text{Denominator: } \frac{d}{dx}[x(1 - e^x)] = (1 - e^x) + x(-e^x) = 1 - e^x - xe^x \]
Thus, we have:
\[ b = \lim_{x \to 0} \frac{2\sin(4x)}{1 - e^x - xe^x} \]

3. **Substituting \( x = 0 \) again**:
\[ 1 - e^0 - 0 = 0 \quad \text{(still \( \frac{0}{0} \))} \]

4. **Apply L'Hôpital's Rule again**:
Differentiate again:
\[ \text{Numerator: } \frac{d}{dx}[2\sin(4x)] = 8\cos(4x) \]
\[ \text{Denominator: } \frac{d}{dx}[1 - e^x - xe^x] = -e^x - (e^x + xe^x) = -2e^x - xe^x \]
Thus, we have:
\[ b = \lim_{x \to 0} \frac{8\cos(4x)}{-2e^x - xe^x} = \frac{8\cdot 1}{-2 - 0} = -4 \]

### Step 3: Evaluate \( c = \lim_{x \to 1} \frac{\sqrt{x} - x}{\ln x} \)

1. **Substituting \( x = 1 \)**:
\[ \sqrt{1} - 1 = 0 \quad \text{and} \quad \ln(1) = 0 \]
This gives us the indeterminate form \( \frac{0}{0} \).

2. **Apply L'Hôpital's Rule**:
Differentiate the numerator and denominator:
\[ \text{Numerator: } \frac{d}{dx}[\sqrt{x} - x] = \frac{1}{2\sqrt{x}} - 1 \]
\[ \text{Denominator: } \frac{d}{dx}[\ln x] = \frac{1}{x} \]
Thus, we have:
\[ c = \lim_{x \to 1} \frac{\frac{1}{2\sqrt{x}} - 1}{\frac{1}{x}} = \lim_{x \to 1} \left(\frac{1 - 2\sqrt{x}}{2\sqrt{x}}\cdot x\right) \]

3. **Substituting \( x = 1 \)**:
\[ c = \frac{1 - 2}{2} = -\frac{1}{2} \]

### Summary of Results
- \( a = -\frac{2}{3} \)
- \( b = -4 \)
- \( c = -\frac{1}{2} \)

### Final Comparison
Now we compare \( a, b, c \):
- \( b < a < c \)
- Therefore, the order is \( b < a < c \).