JEE Main 2023 · 30 January · morning · Numerical answer · +4 / −0
lim x → 0 48/x⁴ ∫0ˣ t³/(t⁶+1) d t is equal to ___________.
Answer and solution
Answer 12
Solution
Maths sir · 3 steps
Top and bottom both go to 0, so differentiate once and the integral sign disappears.
- Spot the form. As x → 0, the integral ∫₀ˣ t³/(t⁶+1) dt → 0 and x⁴ → 0. This is the 0/0 form, so L'Hopital's rule applies.
- Differentiate the integral. By the Leibniz rule, d/dx∫₀ˣ t³/(t⁶+1) dt = x³/(x⁶+1). The derivative of x⁴ is 4x³.
lim(x → 0) (48 · x³/(x⁶+1))/4x³ = lim(x → 0) 12/(x⁶+1)
- So. At x = 0, 12/(0+1) = 12. The limit is 12.
The common slip: 48. Writing the derivative of x⁴ as x³ instead of 4x³ gives 48, not 12.
The idea to keep. When a limit has an integral with variable upper limit over a power of x, use L'Hopital: the derivative of ∫₀ˣ f(t) dt is just f(x).
Answer from the NTA final answer key. Solution by PrepWiser.