JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Limits

Let

f (x) = lim_{y \to 0} \frac{( 1 - c o s ( x y )) t a n ( x y )}{y ^{3}}

. Then the number of solutions of the equation

f (x) = sin x

x \in R

is:

Answer: (C)

Solution

Using

1 - cos (x y) \sim \frac{1}{2} (x y)^{2}

and

tan (x y) \sim x y

y \to 0

f (x) = lim_{y \to 0} \frac{\frac{1}{2} ( x y ) ^{2} \cdot ( x y )}{y ^{3}} = \frac{x ^{3}}{2}

. The equation

f (x) = sin x

becomes

\frac{x ^{3}}{2} = sin x

, i.e.

x^{3} = 2 sin x

, which is odd in

x

. Besides

x = 0

, on

(0, \infty)

the increasing cubic

\frac{x ^{3}}{2}

overtakes

sin x

exactly once (it already exceeds

1 \geq sin x

before

x = \frac{π}{2}

), and by symmetry there is one matching negative root. The number of solutions is

3

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