First slide

Introduction to limits

Question

Let $α and β$ be the distinct roots of ${ax}^{2} + bx + c = 0$ then the value of $lim_{x \to α} \frac{1 - \cos ({ax}^{2} + bx + c)}{(x - α)^{2}}$ is

Easy

A

$\frac{a^{2}}{2} (α - β)^{2}$
B

0
C

$\frac{a^{2}}{2} (α + β)^{2}$
D

$\frac{1}{2} (α - β)^{2}$

Solution

$lim_{x \to α} \frac{1 - \cos ({ax}^{2} + bx + c)}{(x - α)^{2}}$
as $α and β$ are two distinct roots.
$\begin{array}{l} ∴ {ax}^{2} + bx + c = a (x - α) (x - β) \\ i.e. α, β = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \\ = lim_{x \to α} \frac{1 - \cos [(x - α) (x - β) a]}{(x - u)^{2}} \end{array}$
$\begin{array}{l} = lim_{x \to α} \frac{2 \sin^{2} \frac{[(x - α) (x - β) a]}{2}}{(x - α)^{2}} \\ = lim_{x \to α} \frac{2 \sin^{2} \frac{[(x - α) (x - β) a]}{2}}{{[\frac{(x - α) (x - β) a}{2}]}^{2}} a^{2} {(\frac{x - β}{4})}^{2} \\ = lim_{x \to α} \frac{2}{4} a^{2} (x - β)^{2} \\ = \frac{a^{2}}{2} (α - β)^{2} \end{array}$

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