$$f(x) = \mathop {\lim }\limits_{n \to \infty } {{\cos (2\pi x) - {x^{2n}}\sin (x - 1)} \over {1 + {x^{2n + 1}} - {x^{2n}}}}$$

For $$|x| < 1,\,f(x) = \cos 2\pi x$$, continuous function

$$|x| > 1,\,f(x) = \mathop {\lim }\limits_{n \to \infty } {{{1 \over {{x^{2n}}}}\cos 2\pi x - \sin (x - 1)} \over {{1 \over {{x^{2n}}}} + x - 1}}$$

$$ = {{ - \sin (x - 1)} \over {x - 1}}$$, continuous

For $$|x| = 1,\,f(x) = \left\{ {\matrix{ 1 & {\mathrm{if}} & {x = 1} \cr { - (1 + \sin 2)} & {\mathrm{if}} & {x = - 1} \cr } } \right.$$

Now,

$$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = - 1,\,\mathop {\lim }\limits_{x \to {1^ - }} f(x) = 1$$, so discontinuous at $$x = 1$$

$$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = 1,\,\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = - {{\sin 2} \over 2}$$, so discontinuous at $$x = - 1$$

$$\therefore$$ $$f(x)$$ is continuous for all $$x \in R - \{ - 1,1\} $$