The function f: ℝ → ℝ defined by
f(x)=lim(n → ∞) (cos (2 π x)-x²ⁿ sin (x-1))/(1+x²ⁿ⁺¹-x²ⁿ) is continuous for all x in :

Solution

Maths sir · 4 steps

Split by whether |x| is below 1, above 1, or equal to 1, then check the two joints x = ± 1.

1. Behaviour of the power. As n → ∞, x²ⁿ → 0 when |x| < 1 and x²ⁿ → ∞ when |x| > 1. When |x| > 1, divide the numerator and the denominator by x²ⁿ before taking the limit.

   Region	What survives	f(x)
   |x| < 1	the x²ⁿ terms vanish	cos(2π x)
   |x| > 1	-sin(x-1)/(x-1) after dividing by x²ⁿ	-sin(x-1)/(x-1)
   x = 1	(1 - 0)/(1 + 1 - 1)	1
   x = -1	(1 + sin 2)/(1 - 1 - 1)	-(1 + sin 2)

2. Check x = 1. From the left, cos 2π = 1. From the right, -sin(x-1)/(x-1) → -1. The two sides disagree, so f is discontinuous at x = 1.
3. Check x = -1. From the right, cos(-2π) = 1. From the left, -sin(-2)/-2 = -(sin 2)/2. These differ, and f(-1) = -(1 + sin 2) matches neither, so f is discontinuous at x = -1.
4. So. Everywhere else f is a continuous expression, because x - 1 ≠ 0 when |x| > 1. Hence f is continuous on ℝ - {-1, 1}, option (b).

The idea to keep. For a limit with x²ⁿ inside, break the line into |x|<1, |x|>1 and |x|=1. The only places continuity can fail are the boundary points x = ± 1.

Answer from the NTA final answer key. Solution by PrepWiser.