Let - - - R be such that limx - 0x2 sin - xax-sin x-1- Then 6- - - equals

Applying L hospital's rule, lim_x→ 0x^2sin(β x)ax sin x=lim_x→ 0d/dx(x^2sin(β x))d/dx(ax sin x)=lim_x→ 02xsin(β x)+β x^2cos(β x)a cos x=1 ⇒ a cos(0)=0⇒ ...

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Byju's Answer

Standard XII

Mathematics

Expansions to Remove Indeterminate Form

Let α , βε ...

Question

Let $α, β ε R$ be such that $l i m_{x \to 0} \frac{x^{2} s i n (β x)}{a x - s i n x} = 1$ . Then $6 (α + β)$ equals

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Solution

Applying L-hospital's rule,
${lim}_{x \to 0} \frac{x^{2} sin (β x)}{a x - sin x} = {lim}_{x \to 0} \frac{\frac{d}{d x} (x^{2} sin (β x))}{\frac{d}{d x} (a x - sin x)} = {lim}_{x \to 0} \frac{2 x sin (β x) + β x^{2} cos (β x)}{a - cos x} = 1$

$\Rightarrow a - cos (0) = 0 \Rightarrow a = 1$

$\Rightarrow {lim}_{x \to 0} \frac{\frac{d}{d x} (2 x sin (β x) + β x^{2} cos (β x))}{\frac{d}{d x} (1 - cos x)} = {lim}_{x \to 0} \frac{2 sin (β x) + 2 β x cos (β x) + 2 β x cos (β x) - β^{2} x^{2} sin (β x)}{sin x}$

$6 β = 1$
$β = \frac{1}{6}$

So, $6 (α + β) = 6 (1 + \frac{1}{6}) = 7$

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Let α,βεR be such that limx→ 0x2sin(βx)ax−sinx=1. Then 6(α+β) equals

Let $α, β ε R$ be such that $l i m_{x \to 0} \frac{x^{2} s i n (β x)}{a x - s i n x} = 1$ . Then $6 (α + β)$ equals