Skip to the question JEE Main › Mathematics › Limits, Continuity and Differentiability +4 marks−1 if incorrectNumerical pyq
Limits, Continuity and Differentiability: Mathematics | JEE Main lim x → 0 48 x 4 ∫ 0 x t 3 t 6 + 1 d t \lim_{x\to 0} \frac{48}{x^4} \int_0^x \frac{t^3}{t^6+1} dt lim x → 0 x 4 48 ∫ 0 x t 6 + 1 t 3 d t is equal to
No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.
Lock in my answer
Type the value - units or words beside it are fine.
~ ` ! 1 @ 2 # 3 $ 4 % 5 ^ 6 & 7 * 8 ( 9 ) 0 _ - + = Backspace ⌫
Tab Tab q w e r t y u i o p { [ } ] | \
Caps ⇪ a s d f g h j k l : ; " ' Enter ⏎
Shift ⇧ z x c v b n m < , > . ? / Shift ⇧
Alt Alt SPACEBAR SPACEBAR Alt Alt
Choose one answer I’d rather see the answer directly Step-by-step solution View Correct answer
The limit evaluates to 12. Option analysis
Why each option works or fails
Reviewed route
Solution Full solution shortcut
Step Working
01 given Expression: L = lim x → 0 48 x 4 ∫ 0 x t 3 t 6 + 1 d t L = \lim_{x\to 0} \frac{48}{x^4} \int_0^x \frac{t^3}{t^6+1} \, dt L = lim x → 0 x 4 48 ∫ 0 x t 6 + 1 t 3 d t
02 goal Evaluate the limit L L L x → 0 x \to 0 x → 0
03 approach As x → 0 x \to 0 x → 0 ∫ 0 0 t 3 t 6 + 1 d t = 0 \int_0^0 \frac{t^3}{t^6+1} \, dt = 0 ∫ 0 0 t 6 + 1 t 3 d t = 0 x 4 → 0 x^4 \to 0 x 4 → 0 0 0 \frac{0}{0} 0 0
04 execute Differentiate numerator and denominator with respect to x x x d d x ( 48 ∫ 0 x t 3 t 6 + 1 d t ) = 48 ⋅ x 3 x 6 + 1 \frac{d}{dx} \left(48 \int_0^x \frac{t^3}{t^6+1} \, dt\right) = 48 \cdot \frac{x^3}{x^6+1} d x d ( 48 ∫ 0 x t 6 + 1 t 3 d t ) = 48 ⋅ x 6 + 1 x 3 d d x ( x 4 ) = 4 x 3 \frac{d}{dx}(x^4) = 4x^3 d x d ( x 4 ) = 4 x 3 L = lim x → 0 48 ⋅ x 3 x 6 + 1 4 x 3 = lim x → 0 48 4 ( x 6 + 1 ) L = \lim_{x\to 0} \frac{48 \cdot \frac{x^3}{x^6+1}}{4x^3} = \lim_{x\to 0} \frac{48}{4(x^6+1)} L = lim x → 0 4 x 3 48 ⋅ x 6 + 1 x 3 = lim x → 0 4 ( x 6 + 1 ) 48
✓ verify Near t = 0 t = 0 t = 0 t 3 t 6 + 1 ≈ t 3 \frac{t^3}{t^6+1} \approx t^3 t 6 + 1 t 3 ≈ t 3 ∫ 0 x t 3 d t = x 4 4 \int_0^x t^3 \, dt = \frac{x^4}{4} ∫ 0 x t 3 d t = 4 x 4 48 x 4 ( x 4 4 ) = 12 \frac{48}{x^4} \left(\frac{x^4}{4}\right) = 12 x 4 48 ( 4 x 4 ) = 12
Your next move We think you should solve this next ✓ Source and academic review↓
Question type Numerical
Exam relevance JEE Main · Mathematics
Concepts assessed Mathematics
Academic status Reviewed by official_key
Source pyq
Editorial review 9 September 2026 Quick checks
Students also ask Why can we replace the integrand variable t directly with x when differentiating? By the fundamental theorem of calculus (Leibniz rule), d d x ∫ 0 x f ( t ) d t = f ( x ) ⋅ d d x ( x ) − f ( 0 ) ⋅ d d x ( 0 ) = f ( x ) \frac{d}{dx} \int_0^x f(t) \, dt = f(x) \cdot \frac{d}{dx}(x) - f(0) \cdot \frac{d}{dx}(0) = f(x) d x d ∫ 0 x f ( t ) d t = f ( x ) ⋅ d x d ( x ) − f ( 0 ) ⋅ d x d ( 0 ) = f ( x )
Answer The limit evaluates to 12.
Why each option works or fails Step-by-step solution given: Expression: L = lim x → 0 48 x 4 ∫ 0 x t 3 t 6 + 1 d t L = \lim_{x\to 0} \frac{48}{x^4} \int_0^x \frac{t^3}{t^6+1} \, dt L = lim x → 0 x 4 48 ∫ 0 x t 6 + 1 t 3 d t goal: Evaluate the limit L L L x → 0 x \to 0 x → 0 approach: As x → 0 x \to 0 x → 0 ∫ 0 0 t 3 t 6 + 1 d t = 0 \int_0^0 \frac{t^3}{t^6+1} \, dt = 0 ∫ 0 0 t 6 + 1 t 3 d t = 0 x 4 → 0 x^4 \to 0 x 4 → 0 0 0 \frac{0}{0} 0 0 execute: Differentiate numerator and denominator with respect to x x x d d x ( 48 ∫ 0 x t 3 t 6 + 1 d t ) = 48 ⋅ x 3 x 6 + 1 \frac{d}{dx} \left(48 \int_0^x \frac{t^3}{t^6+1} \, dt\right) = 48 \cdot \frac{x^3}{x^6+1} d x d ( 48 ∫ 0 x t 6 + 1 t 3 d t ) = 48 ⋅ x 6 + 1 x 3 d d x ( x 4 ) = 4 x 3 \frac{d}{dx}(x^4) = 4x^3 d x d ( x 4 ) = 4 x 3 L = lim x → 0 48 ⋅ x 3 x 6 + 1 4 x 3 = lim x → 0 48 4 ( x 6 + 1 ) L = \lim_{x\to 0} \frac{48 \cdot \frac{x^3}{x^6+1}}{4x^3} = \lim_{x\to 0} \frac{48}{4(x^6+1)} L = lim x → 0 4 x 3 48 ⋅ x 6 + 1 x 3 = lim x → 0 4 ( x 6 + 1 ) 48 verify: Near t = 0 t = 0 t = 0 t 3 t 6 + 1 ≈ t 3 \frac{t^3}{t^6+1} \approx t^3 t 6 + 1 t 3 ≈ t 3 ∫ 0 x t 3 d t = x 4 4 \int_0^x t^3 \, dt = \frac{x^4}{4} ∫ 0 x t 3 d t = 4 x 4 48 x 4 ( x 4 4 ) = 12 \frac{48}{x^4} \left(\frac{x^4}{4}\right) = 12 x 4 48 ( 4 x 4 ) = 12 Shortcut: When to use it: When the integrand has a simple Maclaurin expansion near the expansion point 0.
quick_kill: Since t → 0 t \to 0 t → 0 t 3 t 6 + 1 = t 3 ( 1 − t 6 + … ) = t 3 + O ( t 9 ) \frac{t^3}{t^6+1} = t^3(1 - t^6 + \dots) = t^3 + O(t^9) t 6 + 1 t 3 = t 3 ( 1 − t 6 + … ) = t 3 + O ( t 9 ) ∫ 0 x t 3 d t = x 4 4 \int_0^x t^3 \, dt = \frac{x^4}{4} ∫ 0 x t 3 d t = 4 x 4 lim x → 0 48 ⋅ x 4 / 4 x 4 = 12 \lim_{x\to 0} 48 \cdot \frac{x^4/4}{x^4} = 12 lim x → 0 48 ⋅ x 4 x 4 /4 = 12