First slide

Introduction to limits

Question

Let $α, β \in R$ be such that $lim_{x \to 0} \frac{x^{2} \sin β x}{α x - \sin x} = 1 .$ Then, $6 (α + β)$ equals

Moderate

A

6
B

7
C

2
D

12

Solution

We find that
$\begin{array}{l} lim_{x \to 0} \frac{x^{2} \sin β x}{α x - \sin x} \\ = lim_{x \to 0} \frac{x (\sin β x)}{α - \sin x} = \frac{0}{α - 1} = 0, if α \neq 1 . \end{array}$
But, it is given that $lim_{x \to 0} \frac{x^{2} \sin β x}{α x - \sin x} = 1 .$ So, $α = 1$
Now,
$lim_{x \to 0} \frac{x^{2} \sin β x}{α x - \sin x} x = 1$
$\Rightarrow lim_{x \to 0} \frac{β x^{3} (\begin{matrix} \sin β x \\ β x \end{matrix})}{x - \sin x} = 1 [∵ α = 1]$
$= lim_{x \to 0} \frac{(\frac{\sin β x}{β x})}{(\frac{x - \sin x}{x^{3}})} = \frac{1}{β}$
$= 6 = \frac{1}{β} \Rightarrow β = \frac{1}{6} [∵ lim_{x \to 0} \frac{x - \sin x}{x^{3}} = \frac{1}{6}]$
Hence, $6 (α + β) = 6 (1 + \frac{1}{6}) = 7$ .

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