LimitsmediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Limits: Let Number Solutions (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question

Let

f(x) = \displaystyle\lim_{y\to0}\dfrac{(1-\cos(xy))\tan(xy)}{y^3}

. The number of solutions of

f(x)=\sin x

x\in\mathbb{R}

, is:

0

2

3

correct

1

Solution

Step 1: Evaluate

f(x)

by substituting

u=xy

f(x) = \lim_{y\to0}\frac{(1-\cos(xy))\tan(xy)}{y^3} = x^3\lim_{u\to0}\frac{(1-\cos u)\tan u}{u^3}

Using the standard limits

\dfrac{1-\cos u}{u^2}\to\dfrac{1}{2}

and

\dfrac{\tan u}{u}\to1

f(x) = x^3\cdot\frac{1}{2}\cdot1 = \frac{x^3}{2}

Step 2: Analyse the equation

h(x) = \dfrac{x^3}{2}-\sin x = 0

Since

h

is an odd function, its positive and negative solutions are symmetric about the origin. Also

h(0)=0

, giving one solution at

x=0

. Step 3: Count solutions for

x>0

h'(x) = \dfrac{3x^2}{2}-\cos x

. Since

h'(0)=-1<0

and

h'(x)\to+\infty

, there exists a unique local minimum at some

x_0>0

. Direct evaluation confirms

h(x_0)<0

, and

h(x)\to+\infty

x\to+\infty

, so exactly one positive zero exists. By odd symmetry, exactly one negative zero also exists. Total solutions:

1+1+1 = 3

. Answer: (3)

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