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Continuity of f(x) = limit of (cos(pi x) - x^(2n) sin(x - 1))/(1 + x^(2n)(x - 1))

The same question, asked 2 times: 28 Jul 2022*, 28 Jan 2026 S2. Every wording below, oldest first.

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  1. Question 1

    28 Jul 2022

    155
    The function f:R→Rf:R \rightarrow R defined by f(x)=lim⁡n→∞cos⁡(2πx)−x2nsin⁡(x−1)1+x2n+1−x2nf(x) =\lim_{n \rightarrow \infty} \frac{\cos(2\pi x) -x^{2n}\sin(x - 1)}{1 +x^{2n + 1}-x^{2n}} is continuous for all xx in

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  2. Question 2

    28 Jan 2026 S2

    55
    Let f(x)=lim⁡θ→0(cos⁡πx−x(2θ)sin⁡(x−1)1+x(2θ)(x−1)),x∈Rf(x) =\lim_{\theta\rightarrow 0} \left( \frac{\cos\pi x-x^{\left( \frac{2}{\theta} \right)}\sin(x- 1)}{1 +x^{\left( \frac{2}{\theta} \right)}(x- 1)} \right),x\in R. Consider the following two statements : (I) f(x)f(x) is discontinous at x=1x = 1. (II) f(x)f(x) is continous at x=−1x = - 1. Then,

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