Question

If $\alpha, \beta$ are the distinct roots of $x^2 + bx + c = 0$, then $\lim_{x \to \beta} \frac{e^{2(x^2 + bx + c)} - 1 - 2(x^2 + bx + c)}{(x - \beta)^2}$ is equal to :

• A. $b^2 - 4c$
• B. $b^2 + 4c$
• C. $2(b^2 + 4c)$
• D. $2(b^2 - 4c)$

Hint:

Using Taylor expansion for $e^z = 1 + z + z^2/2 + \dots$ makes solving limits with $(e^z - 1 - z)$ extremely easy. The leading term is simply $z^2/2$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We use the standard limit $\lim_{t \to 0} \frac{e^t - 1 - t}{t^2} = \frac{1}{2}$.
Step 2: Detailed Explanation:
Let $f(x) = x^2 + bx + c = (x - \alpha)(x - \beta)$.
As $x \to \beta$, $f(x) \to 0$. Let $t = 2f(x)$.
The limit becomes:
\[ \lim_{x \to \beta} \frac{e^t - 1 - t}{t^2} \cdot \frac{t^2}{(x - \beta)^2} \]
\[ = \frac{1}{2} \cdot \lim_{x \to \beta} \frac{[2(x - \alpha)(x - \beta)]^2}{(x - \beta)^2} \]
\[ = \frac{1}{2} \cdot \lim_{x \to \beta} 4(x - \alpha)^2 = 2(\beta - \alpha)^2 \]
We know $(\beta - \alpha)^2 = (\beta + \alpha)^2 - 4\alpha\beta = b^2 - 4c$.
So, the limit is $2(b^2 - 4c)$.
Step 3: Final Answer:
The limit is $2(b^2 - 4c)$.