JEE Main 2022 · 28 July · morning · Numerical answer · +4 / −0
lim(x → 0)(((x+2 cos x)³+2(x+2 cos x)²+3 sin (x+2 cos x))/((x+2)³+2(x+2)²+3 sin (x+2)))^(100/x) is equal to ___________.
Answer and solution
Answer 1
Solution
Maths sir · 5 steps
Top and bottom are one function at inputs that differ by only about x², so the limit is e⁰.
- Same function twice. Let f(t) = t³ + 2t² + 3 sin t, so the base is f(x+2 cos x)/f(x+2). As x → 0 both inputs tend to 2, so the base tends to 1 while the exponent 100/x blows up. This is a 1^∞ form.
- One to infinity rule. For a 1^∞ form, the limit is e^L, where L is the limit of (base - 1) times the exponent.
L = lim(x→ 0) 100/f(x+2)·(f(x+2 cos x)-f(x+2))/x
- Inputs nearly equal. The two inputs differ by 2 cos x - 2 = -4 sin² x/2 ≈ -x². So f(x+2 cos x)-f(x+2) ≈ f'(2) (-x²), where f'(2) = 12 + 8 + 3 cos 2 = 20 + 3 cos 2 is a finite number. Dividing by x leaves about -f'(2) x, which tends to 0.
- Check the denominator. f(2) = 8 + 8 + 3 sin 2 = 16 + 3 sin 2, which is not zero. So L = 100/f(2) × 0 = 0.
- So. The limit is e⁰ = 1.
The idea to keep. In a 1^∞ limit, never stop at 1: the answer is e raised to the limit of (base - 1) times the exponent. Here that product vanished, so the answer came out as 1.
Answer from the NTA final answer key. Solution by PrepWiser.