https://file.aiquickdraw.com/mathbot/file/1727108148582r0o7j6m5.jpg

The problem presented in the image is:

\[
\lim_{x \to 1} \frac{\sin(ax^2 + bx + 1)}{x^2 - 1} = 1
\]

We are asked to find \( a^2 + b \).

### Step-by-Step Solution

1. **Recognize the form**: This is a limit problem involving a sine function. To solve it, we need to simplify the expression, especially since the denominator \(x^2 - 1\) factors into \((x - 1)(x + 1)\), and the limit is as \( x \to 1 \).

2. **Apply the limit**: As \( x \to 1 \), both the numerator and the denominator approach 0. This suggests that L'Hôpital's Rule or Taylor expansion can be useful since it is a \( \frac{0}{0} \) form.

3. **Taylor Expansion (First-Order Approximation)**: Near \(x = 1\), we can approximate \( \sin(z) \) as \( \sin(z) \approx z \) when \( z \) is small. Therefore, we need the expression inside the sine function to also tend toward 0 as \(x \to 1\).

   Evaluate the expression inside the sine function:
   \[
   \lim_{x \to 1} (ax^2 + bx + 1) = a(1)^2 + b(1) + 1 = a + b + 1
   \]

   For the sine approximation to work, we need this expression to tend to 0. Hence:
   \[
   a + b + 1 = 0 \quad \text{or} \quad a + b = -1 \quad \text{(Equation 1)}
   \]

4. **L'Hôpital's Rule**: We differentiate both the numerator and the denominator with respect to \(x\):
   
   \[
   \text{Numerator's derivative: } \frac{d}{dx} \sin(ax^2 + bx + 1) = \cos(ax^2 + bx + 1) \cdot (2ax + b)
   \]
   \[
   \text{Denominator's derivative: } \frac{d}{dx} (x^2 - 1) = 2x
   \]

   At \(x = 1\):
   \[
   \frac{\cos(a + b + 1) \cdot (2a + b)}{2} = 1
   \]
   Since \( \cos(0) = 1 \) (because \( a + b + 1 = 0 \)), the equation becomes:
   \[
   \frac{(2a + b)}{2} = 1 \quad \text{or} \quad 2a + b = 2 \quad \text{(Equation 2)}
   \]

5. **Solve the system of equations**:
   From Equation 1: \( a + b = -1 \)
   
   From Equation 2: \( 2a + b = 2 \)
   
   Subtract Equation 1 from Equation 2:
   \[
   (2a + b) - (a + b) = 2 - (-1)
   \]
   \[
   a = 3
   \]
   
   Substitute \( a = 3 \) into Equation 1:
   \[
   3 + b = -1 \quad \Rightarrow \quad b = -4
   \]

6. **Find \( a^2 + b \)**:
   \[
   a^2 + b = 3^2 + (-4) = 9 - 4 = 5
   \]

Thus, the value of \( a^2 + b \) is **5**.

---

Would you like any further explanation, or details about a specific step?

Here are 5 related questions you might explore:

1. How do you apply L'Hôpital's Rule in different limit cases?
2. What is the Taylor expansion of sine and cosine functions, and how is it useful in limits?
3. Why is the condition \( a + b + 1 = 0 \) necessary in this problem?
4. What are other methods to solve indeterminate forms in limits besides L'Hôpital's Rule?
5. How do you solve systems of linear equations involving two variables?

**Tip**: For problems involving limits, always check if the result is an indeterminate form (like \( \frac{0}{0} \))—it hints at techniques like L'Hôpital's Rule or series expansion.

Jika lim (x -> 1) [sin(ax^2 + bx + 1) / (x^2 - 1)] = 1, maka a^2 + b = ?

Solve the limit lim (x -> 1) [sin(ax^2 + bx + 1) / (x^2 - 1)] = 1 using Taylor expansion and L'Hôpital's Rule to find a^2 + b. This problem applies key concepts from limits, calculus, and trigonometric approximations.

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