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)=lim⁡y→0(1−cos⁡(xy))tan⁡(xy)y3f(x) = \displaystyle\lim_{y\to0}\dfrac{(1-\cos(xy))\tan(xy)}{y^3}. The number of solutions of f(x)=sin⁡xf(x)=\sin x, x∈Rx\in\mathbb{R}, is:
A00
B22
C33correct
D11
Solution
Step 1: Evaluate f(x)f(x) by substituting u=xyu=xy
f(x)=lim⁡y→0(1−cos⁡(xy))tan⁡(xy)y3=x3lim⁡u→0(1−cos⁡u)tan⁡uu3f(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 1−cos⁡uu2→12\dfrac{1-\cos u}{u^2}\to\dfrac{1}{2} and tan⁡uu→1\dfrac{\tan u}{u}\to1:
f(x)=x3⋅12⋅1=x32f(x) = x^3\cdot\frac{1}{2}\cdot1 = \frac{x^3}{2}
 Step 2: Analyse the equation h(x)=x32−sin⁡x=0h(x) = \dfrac{x^3}{2}-\sin x = 0 Since hh is an odd function, its positive and negative solutions are symmetric about the origin. Also h(0)=0h(0)=0, giving one solution at x=0x=0. Step 3: Count solutions for x>0x>0 h′(x)=3x22−cos⁡xh'(x) = \dfrac{3x^2}{2}-\cos x. Since h′(0)=−1<0h'(0)=-1<0 and h′(x)→+∞h'(x)\to+\infty, there exists a unique local minimum at some x0>0x_0>0. Direct evaluation confirms h(x0)<0h(x_0)<0, and h(x)→+∞h(x)\to+\infty as x→+∞x\to+\infty, so exactly one positive zero exists. By odd symmetry, exactly one negative zero also exists. Total solutions: 1+1+1=31+1+1 = 3. Answer: (3)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Limits · easy
lim⁡h→0[1h8+h3−12h]\displaystyle\lim_{h\to 0}\left[\dfrac{1}{h\sqrt[3]{8+h}}-\dfrac{1}{2h}\right] is equal to
Limits · medium
If lim⁡x→0f(x)x2=a\displaystyle\lim_{x\to 0}\dfrac{f(x)}{x^{2}}=a and lim⁡x→0f(1−cos⁡x)g(x)sin⁡2x=b\displaystyle\lim_{x\to 0}\dfrac{f(1-\cos x)}{g(x)\sin^{2}x}=b where bb is not equal to zero then lim⁡x→0g(1−cos⁡2x)x4\displaystyle\lim_{x\to 0}\dfrac{g(1-\cos 2x)}{x^{4}} is
Limits · medium
lim⁡x→∞x3sin⁡(1x)−2x21+3x2\displaystyle\lim_{x\to\infty}\dfrac{x^{3}\sin\left(\dfrac{1}{x}\right)-2x^{2}}{1+3x^{2}} is equal to
Limits · medium
If lim⁡n→∞[(cos⁡kπ4)n−(cos⁡kπ6)n]=0\displaystyle\lim_{n\to\infty}\left[\left(\cos\dfrac{k\pi}{4}\right)^{n}-\left(\cos\dfrac{k\pi}{6}\right)^{n}\right]=0 (where kk is an integer), then which of the following statement(s) is(are) correct?

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath