1. We need to first evaluate the limit

$$f(x)=\lim_{n\to\infty}\frac{\cos(2\pi x)-x^{2n}\sin(x-1)}{1+x^{2n+1}-x^{2n}}.$$

The behavior depends on the value of $|x|$.

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2. Case 1: $|x|<1$

Then

$x^{2n}\to 0, \qquad x^{2n+1}\to 0.$

So the expression becomes

$$f(x)=\frac{\cos(2\pi x)-0}{1+0-0}=\cos(2\pi x).$$

Hence for $|x|<1$,

$f(x)=\cos(2\pi x).$

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3. Case 2: $x=1$

Substitute directly:

$$\frac{\cos(2\pi\cdot 1)-1^{2n}\sin(1-1)}{1+1^{2n+1}-1^{2n}} =\frac{1-0}{1+1-1}=1.$$

So

$f(1)=1.$

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4. Case 3: $x=-1$

Substitute directly:

• Numerator:

$$\cos(-2\pi)-(-1)^{2n}\sin(-2)=1-1\cdot(-\sin 2)=1+\sin 2.$$

• Denominator:

$$1+(-1)^{2n+1}-(-1)^{2n}=1-1-1=-1.$$

Thus

$$f(-1)=\frac{1+\sin 2}{-1}=-(1+\sin 2).$$

So $f(-1)$ exists.

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5. Case 4: $|x|>1$

Now $x^{2n}\to \infty$ in magnitude, and we factor $x^{2n}$ from numerator and denominator:

$$\frac{\cos(2\pi x)-x^{2n}\sin(x-1)}{1+x^{2n+1}-x^{2n}} = \frac{x^{2n}\left(\dfrac{\cos(2\pi x)}{x^{2n}}-\sin(x-1)\right)}{x^{2n}\left(\dfrac{1}{x^{2n}}+x-1\right)}.$$

Therefore,

f(x)=\lim_{n\to\infty} \frac{\dfrac{\cos(2\pi x)}{x^{2n}}-\sin(x-1)}{\dfrac{1}{x^{2n}}+x-1} =rac{-\sin(x-1)}{x-1}, \qquad (|x|>1).

So for $|x|>1$,

$f(x)=\frac{-\sin(x-1)}{x-1}.$

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6. Hence the piecewise form is

$$f(x)= \begin{cases} \cos(2\pi x), & |x|<1,\\[4pt] 1, & x=1,\\[4pt] -(1+\sin2), & x=-1,\\[4pt] \dfrac{-\sin(x-1)}{x-1}, & |x|>1. \end{cases}$$

Now check continuity.

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7. Continuity for interior regions

• On $(-1,1)$, $f(x)=\cos(2\pi x)$, which is continuous.
• On $(1,\infty)$ and $(-\infty,-1)$,

$\frac{-\sin(x-1)}{x-1}$

is continuous wherever $x\ne 1$. In these intervals, $x\ne 1$, so it is continuous.

Thus possible issues are only at $x=1$ and $x=-1$.

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8. Check continuity at $x=1$

Left-hand limit ($x\to 1^-$): since $|x|<1$ nearby,

$$\lim_{x\to 1^-} f(x)=\lim_{x\to 1^-}\cos(2\pi x)=\cos(2\pi)=1.$$

Right-hand limit ($x\to 1^+$): since $x>1$ nearby,

$$\lim_{x\to 1^+} f(x)=\lim_{x\to 1^+}\frac{-\sin(x-1)}{x-1}=-1$$

because

$\lim_{t\to 0}\frac{\sin t}{t}=1.$

Since

$1\ne -1,$

$f$ is discontinuous at $x=1$.

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9. Check continuity at $x=-1$

Left-hand limit ($x\to -1^-$): use $|x|>1$ formula,

$$\lim_{x\to -1^-} f(x)=\lim_{x\to -1^-}\frac{-\sin(x-1)}{x-1} =\frac{-\sin(-2)}{-2}=-\frac{\sin 2}{2}.$$

Right-hand limit ($x\to -1^+$): use $|x|<1$ formula,

$$\lim_{x\to -1^+} f(x)=\cos(-2\pi)=1.$$

These are not equal, so $f$ is discontinuous at $x=-1$.

Also they do not match $f(-1)=-(1+\sin2)$.

Hence $f$ is discontinuous at $x=-1$ also.

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10. Therefore $f$ is continuous for all real $x$ except at $x=\pm 1$.

So the set is

$\mathbb{R}-\{-1,1\}.$

Thus the correct option is B.