Question

If \(\underset{x\rightarrow 1}{lim}\frac{x^3+ax^2+bx+c}{x^2-2x+1} = 2026\) then the value of \(a-c\) is...

• A. \(2\)
• B. \(1\)
• C. \(-1\)
• D. \(-2\)

Hint:

For a finite limit the numerator needs a double root at x = 1, so write it as (x-1)^2 (x + k).

Answer 1

The Correct Option is D

Step 1: Apply L'Hopital twice
Let $N(x)$ be the numerator. Since the limit is finite, $N(1) = 0$ and $N'(1) = 0$, so $1 + a + b + c = 0$ and $3 + 2a + b = 0$.

Step 2: Second derivative
Applying the rule twice, the limit is $\dfrac{N''(1)}{2} = \dfrac{6 + 2a}{2} = 3 + a = 2026$, so $a = 2023$.

Step 3: Find the rest
From $3 + 2a + b = 0$: $b = -4049$. Then $c = -1 - a - b = -1 - 2023 + 4049 = 2025$.

Step 4: Result
$a - c = 2023 - 2025 = -2$.

Final Answer:
a - c equals -2. This is option (D). \[ \boxed{\text{(D) }-2} \]