Question

The function \(f : R→R\) defined by \(f(x)=lim_{n→∞}\frac{cos2πx-x^{2n}sinx-1}{1+x^{2n+1}-x^{2n}}\) is continuous for all x in

• A. R-{-1}
• B. R-{-1,1}
• C. R-{1}
• D. R-{0}

Answer 1

The Correct Option is B

To determine the points of continuity for the function f(x)=lim_{n→∞}\frac{cos2πx-x^{2n}sinx-1}{1+x^{2n+1}-x^{2n}}, we need to consider the behavior of the numerator and denominator as n approaches infinity for different values of x.

1. Consider the cases for the value of x. If |x|<1:
   • The term x^{2n} and x^{2n+1} tend to zero as n \to \infty.
   • The function simplifies to f(x)=\frac{\cos2\pi x - \sin x - 1}{1}, which is continuous for x \neq -1, 1, as the sine and cosine functions are continuous everywhere.
2. For x = 1:
   • The numerator simplifies to \cos 2\pi \cdot 1 - (1)^{2n}\sin 1 - 1 = 1 - \sin 1 - 1 = -\sin 1, a constant.
   • The denominator becomes 1 + 1^{2n+1} - 1^{2n} = 1, so f(x) tends to a non-zero value. Evaluating f(x) gives -\sin 1, indicating a continuous value at x = 1 itself.
3. For x = -1:
   • The numerator is \cos(-2\pi)-(-1)^{2n}\sin(-1)-1 = 1 + (-1)^n \sin 1 - 1.
   • The denominator will fluctuate between 0 and 2 depending on whether n is odd or even, leading to inconsistency in limits, indicating a discontinuity at x = -1.

Therefore, the function is continuous for all x except -1 and 1. Thus, the continuity is defined on the set \mathbb{R} - \{-1, 1\}.