If `Lim_(xto1) (x^(2)-ax+b)/(x-1)=5` then the value of `a+b` is

Options

A. 1
B. `-5`
C. 4
D. `-7`

Answer

To solve the limit problem given, we need to find the values of \( a \) and \( b \) such that:

\[ \lim_{x \to 1} \frac{x^2 - ax + b}{x - 1} = 5 \]

### Step 1: Substitute \( x = 1 \) in the limit expression

When we substitute \( x = 1 \):

\[ \frac{1^2 - a(1) + b}{1 - 1} = \frac{1 - a + b}{0} \]

This results in an indeterminate form \( \frac{0}{0} \). For the limit to exist, the numerator must also equal zero when \( x = 1 \):

\[ 1 - a + b = 0 \]

### Step 2: Rearranging the equation

From the equation \( 1 - a + b = 0 \), we can express \( b \) in terms of \( a \):

\[ b = a - 1 \quad \text{(Equation 1)} \]

### Step 3: Apply L'Hôpital's Rule

Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. We differentiate the numerator and the denominator separately:

- The derivative of the numerator \( x^2 - ax + b \) is \( 2x - a \).
- The derivative of the denominator \( x - 1 \) is \( 1 \).

Thus, we have:

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

### Step 4: Evaluate the limit

Now substituting \( x = 1 \):

\[ \lim_{x \to 1} (2x - a) = 2 - a \]

According to the problem, this limit equals 5:

\[ 2 - a = 5 \]

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

Rearranging the equation gives:

\[ -a = 5 - 2 \implies -a = 3 \implies a = -3 \]

### Step 6: Substitute \( a \) back to find \( b \)

Using Equation 1 \( b = a - 1 \):

\[ b = -3 - 1 = -4 \]

### Step 7: Find \( a + b \)

Now we can find \( a + b \):

\[ a + b = -3 + (-4) = -7 \]

Thus, the final answer is:

\[ \boxed{-7} \]