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JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Limits XII

Let

α, β \in R

be such that

lim_{x \to 0} \frac{x ^{2} s i n ( β x )}{α x - s i n x} = 1

. Then

6 (α + β)

equals

Answer: 7

Solution

Rewrite the limit:

x \to 0 lim \frac{x ^{2} sin ( β x )}{α x - sin x} = x \to 0 lim \frac{x ^{2} \cdot β x \cdot \frac{s i n ( β x )}{β x}}{α x - sin x} = x \to 0 lim \frac{β x ^{3}}{α x - sin x}

Expanding

sin x = x - \frac{x ^{3}}{6} + O (x^{5})

:

α x - sin x = α x - x + \frac{x ^{3}}{6} + O (x^{5}) = (α - 1) x + \frac{x ^{3}}{6} + O (x^{5})

For the limit to exist and equal 1, we need

α - 1 = 0

, so

α = 1

.

Then:

x \to 0 lim \frac{β x ^{3}}{\frac{x ^{3}}{6}} = 6 β = 1 \Rightarrow β = \frac{1}{6}

Therefore:

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

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Integer answer

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