First slide

Introduction to limits

Question

The value of $lim_{x \to \frac{1}{α}} \frac{\sin ({cx}^{2} + bx + a)}{xα - 1}$ (where $α$ and $β$ are the roots ${ax}^{2} + bx + c = 0$ ) is

Moderate

A

$c (\frac{β - α}{αβ})$
B

$(\frac{β - α}{α})$
C

$\frac{1}{α}$
D

$\frac{c}{α} (\frac{1}{α} - \frac{1}{β})$

Solution

$∵ Roots of {ax}^{2} + bx + c = 0 are α and β$
so roots of ${cx}^{2} + bx + a = 0 are \frac{1}{α} and \frac{1}{β}$
$\begin{array}{l} ∴ {cx}^{2} + bx + a = c (x - \frac{1}{α}) (x - \frac{1}{β}) \\ Now, lim_{x \to \frac{1}{α}} \frac{\sin [c (x - \frac{1}{α}) (x - \frac{1}{β})]}{αc (x - \frac{1}{α}) (x - \frac{1}{β})} \cdot c (x - \frac{1}{β}) \end{array}$
$\begin{matrix} = \frac{c}{α} lim_{x \to \frac{1}{α}} [\frac{\sin \{c (x - \frac{1}{α}) (x - \frac{1}{β})\}}{c (x - \frac{1}{α}) (x - \frac{1}{β})}] \cdot lim_{x \to \frac{1}{α}} (x - \frac{1}{β}) \\ = \frac{c}{α} (\frac{1}{α} - \frac{1}{β}) \end{matrix}$

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