If `lim_(xrarr1)(ax^2+bx+c)/((x-1)^2)=2`, then `lim_(xrarr1)((x-a)(x-b)(x-c))/(x+1)` , is

Options

A. 2
B. `(5)/(2)`
C. `(7)/(2)`
D. 5

Explanation

To solve the problem step by step, we start with the given limit:

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

### Step 1: Identify the form of the limit
When we substitute \(x = 1\) in the denominator, we find that \((1 - 1)^2 = 0\). We also need to evaluate the numerator:

\[ a(1)^2 + b(1) + c = a + b + c \]

For the limit to be in the indeterminate form \( \frac{0}{0} \), we require:

\[ a + b + c = 0 \]

### Step 2: Apply L'Hôpital's Rule
Since we have established that both the numerator and denominator approach 0 as \(x\) approaches 1, we can apply L'Hôpital's Rule. We 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 we rewrite the limit:

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

### Step 3: Evaluate the new limit
Substituting \(x = 1\) gives:

\[ \text{Numerator: } 2a(1) + b = 2a + b \]
\[ \text{Denominator: } 2(1-1) = 0 \]

Again, we have a \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule again:

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

Now we have:

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

### Step 4: Set the limit equal to 2
Since we know this limit equals 2:

\[ a = 2 \]

### Step 5: Find \(b\) and \(c\)
Using the condition \(a + b + c = 0\):

\[ 2 + b + c = 0 \implies b + c = -2 \quad (1) \]

Now using \(2a + b = 0\):

\[ 2(2) + b = 0 \implies 4 + b = 0 \implies b = -4 \quad (2) \]

Substituting \(b = -4\) into equation (1):

\[ -4 + c = -2 \implies c = 2 \]

### Step 6: Values of \(a\), \(b\), and \(c\)
We have:

\[ a = 2, \quad b = -4, \quad c = 2 \]

### Step 7: Evaluate the second limit
Now we need to evaluate:

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

Substituting the values of \(a\), \(b\), and \(c\):

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

### Step 8: Substitute \(x = 1\)
Now substituting \(x = 1\):

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

### Final Answer
Thus, the limit evaluates to:

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