JEE Main 2026 — Mathematics Calculus

Let $\displaystyle L = \lim_{x \to 0} \dfrac{x^2\sin^2 x}{x^2 - \sin^2 x}$L=x→0lim​x2−sin2xx2sin2x​

Dividing the numerator and the denominator by $x^4$x4, we get:

$$L = \lim_{x \to 0} \dfrac{\left(\dfrac{\sin x}{x}\right)^2}{\dfrac{x^2 - \sin^2 x}{x^4}}$$L=x→0lim​x4x2−sin2x​(xsinx​)2​

Factorizing the denominator:

$$L = \lim_{x \to 0} \dfrac{\left(\dfrac{\sin x}{x}\right)^2}{\left(\dfrac{x - \sin x}{x^3}\right) \left(\dfrac{x + \sin x}{x}\right)}$$L=x→0lim​(x3x−sinx​)(xx+sinx​)(xsinx​)2​

Using the standard limits:

$$\lim_{x \to 0} \dfrac{\sin x}{x} = 1$$x→0lim​xsinx​=1

$\displaystyle \lim_{x \to 0} \dfrac{x - \sin x}{x^3} = \dfrac{1}{6}$x→0lim​x3x−sinx​=61​ (using Taylor series expansion $\sin x = x - \dfrac{x^3}{3!} + \dots$sinx=x−3!x3​+…)

$$\lim_{x \to 0} \dfrac{x + \sin x}{x} = \lim_{x \to 0} \left(1 + \dfrac{\sin x}{x}\right) = 1 + 1 = 2$$x→0lim​xx+sinx​=x→0lim​(1+xsinx​)=1+1=2

Substituting these limits into the expression:

$$L = \dfrac{1^2}{\left(\dfrac{1}{6}\right) \times 2} = \dfrac{1}{\dfrac{1}{3}} = 3$$L=(61​)×212​=31​1​=3

Answer: $3$3