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 limits \( a \), \( b \), and \( c \) given in the question, we will evaluate each limit step by step.

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

1. **Use the Taylor series expansion for \(\cos(2x)\)**:
\[ \cos(2x) \approx 1 - 2x^2 \quad \text{as } x \to 0 \]
Therefore,
\[ \ln(\cos(2x)) \approx \ln(1 - 2x^2) \approx -2x^2 \quad \text{(using } \ln(1 + u) \approx u \text{ for small } u\text{)} \]

2. **Substituting back into the limit**:
\[ a = \lim_{x \to 0} \frac{-2x^2}{3x^2} = \lim_{x \to 0} \frac{-2}{3} = -\frac{2}{3} \]

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

1. **Use the Taylor series for \(\sin(2x)\)**:
\[ \sin(2x) \approx 2x \quad \text{as } x \to 0 \]
Therefore,
\[ \sin^2(2x) \approx (2x)^2 = 4x^2 \]

2. **Use the Taylor series for \( e^x \)**:
\[ e^x \approx 1 + x \quad \text{as } x \to 0 \implies 1 - e^x \approx -x \]

3. **Substituting back into the limit**:
\[ b = \lim_{x \to 0} \frac{4x^2}{x(-x)} = \lim_{x \to 0} \frac{4x^2}{-x^2} = \lim_{x \to 0} -4 = 4 \]

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

1. **Rewrite \(\sqrt{x} - x\)**:
\[ \sqrt{x} - x = \frac{(\sqrt{x} - x)(\sqrt{x} + x)}{\sqrt{x} + x} = \frac{x - x^2}{\sqrt{x} + x} = \frac{x(1 - x)}{\sqrt{x} + x} \]

2. **Now substitute into the limit**:
\[ c = \lim_{x \to 1} \frac{x(1 - x)}{\ln x (\sqrt{x} + x)} \]

3. **As \( x \to 1\)**, both the numerator and denominator approach 0, so we apply L'Hôpital's Rule:
- Differentiate the numerator: \( \frac{d}{dx}(x(1 - x)) = 1 - 2x \)
- Differentiate the denominator: \( \frac{d}{dx}(\ln x(\sqrt{x} + x)) = \frac{1}{x}(\sqrt{x} + x) + \ln x \cdot \frac{1}{2\sqrt{x}} + \ln x \)

4. **Evaluate the limit using L'Hôpital's Rule**:
\[ c = \lim_{x \to 1} \frac{1 - 2x}{\frac{1}{x}(\sqrt{x} + x) + \ln x \cdot \frac{1}{2\sqrt{x}} + \ln x} \]
After evaluating, we find:
\[ c = -\frac{1}{2} \]

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

### Final Comparison
Now we compare \( a, b, c \):
- \( a < c < b \)

### Conclusion
Thus, the values of \( a, b, c \) satisfy \( a < c < b \).