To find `lim_(x to 0) (1+(x^2+3x))^(1//sinx)`

Answer

To find the limit \( \lim_{x \to 0} (1 + (x^2 + 3x))^{\frac{1}{\sin x}} \), we can follow these steps:

### Step 1: Identify the Form
First, we substitute \( x = 0 \) into the expression to see what form we get:
\[ 1 + (0^2 + 3 \cdot 0) = 1 + 0 = 1 \]
The exponent becomes:
\[ \frac{1}{\sin(0)} = \frac{1}{0} \text{ (undefined)} \]
Thus, we have the indeterminate form \( 1^{\infty} \).

### Step 2: Rewrite the Expression
Since we have the indeterminate form \( 1^{\infty} \), we can rewrite the limit using the exponential function:
\[ \lim_{x \to 0} (1 + (x^2 + 3x))^{\frac{1}{\sin x}} = e^{\lim_{x \to 0} \left( (1 + (x^2 + 3x) - 1) \cdot \frac{1}{\sin x} \right)} \]
This simplifies to:
\[ e^{\lim_{x \to 0} \frac{x^2 + 3x}{\sin x}} \]

### Step 3: Analyze the New Limit
Now we need to evaluate:
\[ \lim_{x \to 0} \frac{x^2 + 3x}{\sin x} \]
Substituting \( x = 0 \) gives us:
\[ \frac{0^2 + 3 \cdot 0}{\sin(0)} = \frac{0}{0} \text{ (indeterminate)} \]
Since we have \( \frac{0}{0} \), we can apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule
Taking the derivative of the numerator and the denominator:
- Derivative of the numerator \( x^2 + 3x \) is \( 2x + 3 \).
- Derivative of the denominator \( \sin x \) is \( \cos x \).

Now we have:
\[ \lim_{x \to 0} \frac{2x + 3}{\cos x} \]

### Step 5: Evaluate the New Limit
Substituting \( x = 0 \):
\[ \frac{2 \cdot 0 + 3}{\cos(0)} = \frac{3}{1} = 3 \]

### Step 6: Final Result
Now substituting back into the exponential expression:
\[ e^{\lim_{x \to 0} \frac{x^2 + 3x}{\sin x}} = e^3 \]

Thus, the limit is:
\[ \lim_{x \to 0} (1 + (x^2 + 3x))^{\frac{1}{\sin x}} = e^3 \]

### Summary
The final answer is:
\[ \boxed{e^3} \]