If `Delta(x)=|(e^(x^2),log(1+x)),(tanx,sinx)|`, then `{:(Lt),(xrarr0):}(Delta(x))/x=`

Options

A. `-1`
B. 0
C. 1
D. none

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{x \to 0} \frac{\Delta(x)}{x} \]

where

\[ \Delta(x) = \begin{vmatrix} e^{x^2} & \log(1+x) \\ \tan x & \sin x \end{vmatrix} \]

### Step 1: Calculate the Determinant

The determinant \(\Delta(x)\) can be calculated as follows:

\[ \Delta(x) = e^{x^2} \cdot \sin x - \log(1+x) \cdot \tan x \]

### Step 2: Substitute the Limit

Now we substitute this into our limit:

\[ \lim_{x \to 0} \frac{\Delta(x)}{x} = \lim_{x \to 0} \frac{e^{x^2} \sin x - \log(1+x) \tan x}{x} \]

### Step 3: Use L'Hôpital's Rule

Since both the numerator and denominator approach 0 as \(x \to 0\), we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{e^{x^2} \sin x - \log(1+x) \tan x}{x} = \lim_{x \to 0} \frac{d}{dx}(e^{x^2} \sin x - \log(1+x) \tan x) \]

### Step 4: Differentiate the Numerator

We need to differentiate the numerator:

1. For \(e^{x^2} \sin x\):
- Using the product rule:
\[ \frac{d}{dx}(e^{x^2} \sin x) = e^{x^2} \cdot 2x \sin x + e^{x^2} \cos x \]

2. For \(-\log(1+x) \tan x\):
- Using the product rule:
\[ \frac{d}{dx}(-\log(1+x) \tan x) = -\frac{1}{1+x} \tan x - \log(1+x) \sec^2 x \]

So, we have:

\[ \frac{d}{dx}(e^{x^2} \sin x - \log(1+x) \tan x) = e^{x^2} \cdot 2x \sin x + e^{x^2} \cos x - \frac{1}{1+x} \tan x - \log(1+x) \sec^2 x \]

### Step 5: Evaluate the Limit

Now we evaluate the limit as \(x \to 0\):

1. \(e^{x^2} \to e^0 = 1\)
2. \(\sin x \to 0\)
3. \(\cos x \to 1\)
4. \(\tan x \to 0\)
5. \(\sec^2 x \to 1\)
6. \(\log(1+x) \to 0\)

Thus, substituting these values:

\[ \lim_{x \to 0} \left( 1 \cdot 2 \cdot 0 + 1 \cdot 1 - 0 - 0 \right) = 1 \]

### Step 6: Final Limit Calculation

Thus, we have:

\[ \lim_{x \to 0} \frac{\Delta(x)}{x} = 1 \]

### Conclusion

The final answer is:

\[ \boxed{1} \]