the value of `lim_(x->0){(cosx)^(1/(sin^2x))+(sin2x+2tan^-13x+3x^2)/(ln(1+3x+sin^2x)+xe^x)}`

Options

A. `sqrte+3/2`
B. `(1)/(sqrte)+ 3/2`
C. `sqrte +2`
D. `(1)/(sqrte) +2`

Answer

To solve the limit \( \lim_{x \to 0} \left( \cos x^{\frac{1}{\sin^2 x}} + \frac{\sin 2x + 2\tan^{-1}(3x) + 3x^2}{\ln(1 + 3x + \sin^2 x) + xe^x} \right) \), we will break it down into two parts: \( L_1 \) and \( L_2 \).

### Step 1: Evaluating \( L_1 = \lim_{x \to 0} \cos x^{\frac{1}{\sin^2 x}} \)

1. **Substituting \( x = 0 \)**:
\[ \cos(0)^{\frac{1}{\sin^2(0)}} = 1^{\infty} \]
This is an indeterminate form, so we will take the logarithm.

2. **Taking the logarithm**:
\[ \ln L_1 = \lim_{x \to 0} \frac{\ln(\cos x)}{\sin^2 x} \]

3. **Applying L'Hôpital's Rule**:
Since both the numerator and denominator approach 0 as \( x \to 0 \), we can apply L'Hôpital's Rule:
\[ \ln L_1 = \lim_{x \to 0} \frac{-\sin x}{\sin^2 x \cdot \cos x} = \lim_{x \to 0} \frac{-1}{2\cos x} = -\frac{1}{2} \]

4. **Exponentiating to find \( L_1 \)**:
\[ L_1 = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \]

### Step 2: Evaluating \( L_2 = \lim_{x \to 0} \frac{\sin 2x + 2\tan^{-1}(3x) + 3x^2}{\ln(1 + 3x + \sin^2 x) + xe^x} \)

1. **Substituting \( x = 0 \)**:
\[ \sin(0) + 2\tan^{-1}(0) + 3(0)^2 = 0 \]
and
\[ \ln(1 + 0 + 0) + 0 = 0 \]
This is also an indeterminate form \( \frac{0}{0} \).

2. **Applying L'Hôpital's Rule**:
Differentiate the numerator and denominator:
- Numerator: \( 2\cos(2x) + \frac{6}{1 + (3x)^2} + 6x \)
- Denominator: \( \frac{3}{1 + 3x + \sin^2 x} + e^x + xe^x \)

3. **Evaluating the limit**:
Substitute \( x = 0 \):
\[ \text{Numerator: } 2 + 6 + 0 = 8 \]
\[ \text{Denominator: } 3 + 1 + 0 = 4 \]
Thus,
\[ L_2 = \frac{8}{4} = 2 \]

### Final Step: Combining \( L_1 \) and \( L_2 \)

Now, we combine both limits:
\[ \lim_{x \to 0} \left( L_1 + L_2 \right) = \frac{1}{\sqrt{e}} + 2 \]

### Conclusion

The final value of the limit is:
\[ \lim_{x \to 0} \left( \cos x^{\frac{1}{\sin^2 x}} + \frac{\sin 2x + 2\tan^{-1}(3x) + 3x^2}{\ln(1 + 3x + \sin^2 x) + xe^x} \right) = 2 + \frac{1}{\sqrt{e}} \]