The function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by

$$f(x)=\lim\limits_{n \rightarrow \infty} \frac{\cos (2 \pi x)-x^{2 n} \sin (x-1)}{1+x^{2 n+1}-x^{2 n}}$$ is continuous for all x in :

<p>$$f(x) = \mathop {\lim }\limits_{n \to \infty } {{\cos (2\pi x) - {x^{2n}}\sin (x - 1)} \over {1 + {x^{2n + 1}} - {x^{2n}}}}$$</p> <p>For $|x| < 1,\,f(x) = \cos 2\pi x$, continuous function</p> <p>$$|x| > 1,\,f(x) = \mathop {\lim }\limits_{n \to \infty } {{{1 \over {{x^{2n}}}}\cos 2\pi x - \sin (x - 1)} \over {{1 \over {{x^{2n}}}} + x - 1}}$$</p> <p>$= {{ - \sin (x - 1)} \over {x - 1}}$, continuous</p> <p>For $$|x| = 1,\,f(x) = \left\{ {\matrix{ 1 & {\mathrm{if}} & {x = 1} \cr { - (1 + \sin 2)} & {\mathrm{if}} & {x = - 1} \cr } } \right.$$</p> <p>Now,</p> <p>$$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = - 1,\,\mathop {\lim }\limits_{x \to {1^ - }} f(x) = 1$$, so discontinuous at $x = 1$</p> <p>$$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = 1,\,\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = - {{\sin 2} \over 2}$$, so discontinuous at $x = - 1$</p> <p>$\therefore$ $f(x)$ is continuous for all $x \in R - \{ - 1,1\}$</p>