Let `a= lim_(x->0)ln(cos2x)/(3x^2), b=lim_(x->0)(sin^(2)2x)/(x(1-e^x)), c=lim_(x->1)(sqrt(x)-x)/lnx`

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 limits \( a \), \( b \), and \( c \) step by step, we will use L'Hôpital's rule where necessary.

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

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

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

3. **Substituting \( x = 0 \)** again gives \( \frac{0}{0} \), so apply L'Hôpital's Rule again:
\[ \frac{d}{dx}[-\tan(2x)] = -2\sec^2(2x) \]
\[ a = \lim_{x \to 0} \frac{-2\sec^2(2x)}{3} = \frac{-2}{3} \]

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

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

2. **Apply L'Hôpital's Rule**:
Rewrite the limit:
\[ b = \lim_{x \to 0} \frac{4\sin(2x)\cos(2x)}{1 - e^x + xe^x} \]

3. **Differentiate**:
\[ \text{Numerator: } \frac{d}{dx}[4\sin(2x)\cos(2x)] = 8\cos(2x)\cos(2x) - 8\sin(2x)\sin(2x) = 8\cos(2x)\cos(2x) - 8\sin^2(2x) \]
\[ \text{Denominator: } \frac{d}{dx}[1 - e^x + xe^x] = -e^x + e^x + xe^x = xe^x \]
So,
\[ b = \lim_{x \to 0} \frac{8\cos^2(2x)}{e^x} = \frac{8}{1} = 8 \]

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

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

2. **Apply L'Hôpital's Rule**:
Differentiate:
\[ \text{Numerator: } \frac{1}{2\sqrt{x}} - 1 \]
\[ \text{Denominator: } \frac{1}{x} \]
So,
\[ c = \lim_{x \to 1} \frac{\frac{1}{2\sqrt{x}} - 1}{\frac{1}{x}} = \lim_{x \to 1} \frac{x(\frac{1}{2\sqrt{x}} - 1)}{1} = \frac{1}{2} - 1 = -\frac{1}{2} \]

### Summary of Results
- \( a = -\frac{2}{3} \)
- \( b = 8 \)
- \( c = -\frac{1}{2} \)

### Final Comparison
From the results:
- \( c < a < b \)