AIEEE 2005 (Offline)Question 1
Let $\alpha$ and $\beta$ be the distinct roots of $ax^{2}+bx+c=0$, then $\lim_{x\to\alpha}\frac{1-\cos\left(ax^{2}+bx+c\right)}{(x-\alpha)^{2}}$ is equal to
Limits & Continuity
Solution
Concept. Write the quadratic using its roots so that $(x-\alpha)$ appears. Formulas used. (1) $ax^2+bx+c=a(x-\alpha)(x-\beta)$(2) $\lim_{u\to0}\frac{1-\cos u}{u^2}=\frac12$Step 1 - Substitute (1). $u=a(x-\alpha)(x-\beta)\to0$ as $x\to\alpha$. Step 2 - Split. $$\frac{1-\cos u}{(x-\alpha)^2}=\frac{1-\cos u}{u^2}\cdot a^2(x-\beta)^2$$ Step 3 - Limit (2). $\frac12\cdot a^2(\alpha-\beta)^2$. Answer: A