The function f: R → R defined by f(x) = _ n → ∞ (2 ) - x ^ 2n (x -...

Given $f\left(x\right) = \lim_{n \to \infty}\dfrac{\cos 2\pix - {x}^{2n}\sin \left(x - 1\right)}{1 + {x}^{2n + 1} - {x}^{2n}}$

$= \lim_{n \to \infty}\dfrac{\cos 2\pix - {\left({x}^{2}\right)}^{n}\sin \left(x - 1\right)}{1 + x{\left({x}^{2}\right)}^{n} - {\left({x}^{2}\right)}^{n}}$

Now for $- 1 < x < 1,$ as $0 < {x}^{2} < 1 ⇒ \lim_{n \to \infty}{x}^{2n} \to 0$

i.e. $f\left(x\right) = \cos 2\pix$

Now again rewriting $f\left(x\right) = \lim_{n \to \infty}\dfrac{{\left({x}^{2}\right)}^{- n}\cos 2\pix - \sin \left(x - 1\right)}{{\left({x}^{2}\right)}^{- n} + x - 1}$

For $x > 1\text{or}x < - 1,$ $\lim_{n \to \infty}{x}^{- 2n} \to 0$

i.e. $f\left(x\right) = - \dfrac{\sin \left(x - 1\right)}{x - 1}$

For x = \pm 1, f\left(x\right) = \left\{\begin{cases}1 & ifx = 1 \\ \dfrac{\left(1 + \sin 2\right)}{- 1} & ifx = - 1\end{cases}\right)</p><p>i.e.\lim_{x \to {1}^{+}}f\left(x\right) = - 1, \lim_{x \to {1}^{-}}f\left(x\right) = 1$</p><p>So$f$is discontinuous at$x = 1$</p><p>$\lim_{x \to - {1}^{+}}f\left(x\right) = 1, \lim_{x \to - {1}^{-}}f\left(x\right) = - \dfrac{\sin 2}{2}$</p><p>So$f\left(x\right)$is discontinuous at$x = - 1