If `underset(x to 1)lim (ax^(2)+bx+c)/((x-1)^(2))=2" then "underset(x to 1)lim ((x-a)(x-b)(x-c))/((x+1))=`

Options

A. 2
B. `3/2`
C. 4
D. `5/2`

Answer

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

\[ \lim_{x \to 1} \frac{(x-a)(x-b)(x-c)}{(x+1)} \]

given that

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

### Step 1: Analyze the given limit condition

The limit condition indicates that when \( x = 1 \), both the numerator and denominator approach 0, thus forming a \( \frac{0}{0} \) indeterminate form. This means we can apply L'Hôpital's Rule.

### Step 2: Set up the equations from the limit condition

1. **First Equation**: Since \( ax^2 + bx + c \) must equal 0 when \( x = 1 \):

\[ a(1)^2 + b(1) + c = 0 \implies a + b + c = 0 \quad \text{(Equation 1)} \]

2. **Differentiate using L'Hôpital's Rule**:

Differentiate the numerator and denominator:

\[ \text{Numerator: } \frac{d}{dx}(ax^2 + bx + c) = 2ax + b \]
\[ \text{Denominator: } \frac{d}{dx}((x-1)^2) = 2(x-1) \]

Now, applying L'Hôpital's Rule:

\[ \lim_{x \to 1} \frac{2ax + b}{2(x-1)} \]

Again, substituting \( x = 1 \) gives \( 0 \) in the denominator, so we apply L'Hôpital's Rule again.

3. **Second Equation**: Differentiate again:

\[ \text{Numerator: } \frac{d}{dx}(2ax + b) = 2a \]
\[ \text{Denominator: } \frac{d}{dx}(2(x-1)) = 2 \]

Thus, we have:

\[ \lim_{x \to 1} \frac{2a}{2} = a \]

Setting this equal to 2 (from the limit condition):

\[ a = 2 \quad \text{(Equation 2)} \]

### Step 3: Solve for \( b \) and \( c \)

Substituting \( a = 2 \) into Equation 1:

\[ 2 + b + c = 0 \implies b + c = -2 \quad \text{(Equation 3)} \]

### Step 4: Use the limit to find \( b \)

Now, we need to ensure the limit condition holds true. We already found \( a = 2 \). We also need to ensure that the first derivative at \( x = 1 \) is 0:

From the first derivative:

\[ 2(2) + b = 0 \implies 4 + b = 0 \implies b = -4 \quad \text{(Equation 4)} \]

### Step 5: Solve for \( c \)

Substituting \( b = -4 \) into Equation 3:

\[ -4 + c = -2 \implies c = 2 \quad \text{(Equation 5)} \]

### Step 6: Substitute \( a, b, c \) into the limit expression

Now we have \( a = 2, b = -4, c = 2 \). We substitute these values into the limit we want to evaluate:

\[ \lim_{x \to 1} \frac{(x - 2)(x + 4)(x - 2)}{(x + 1)} \]

### Step 7: Evaluate the limit

Substituting \( x = 1 \):

\[ = \frac{(1 - 2)(1 + 4)(1 - 2)}{(1 + 1)} = \frac{(-1)(5)(-1)}{2} = \frac{5}{2} \]

Thus, the final answer is:

\[ \boxed{\frac{5}{2}} \]