The function f: R arrow R defined by f(x)= displaystyle lim n arrow ∞ ( cos (2 π x)-x2 n sin (x-1)/1+x2 n+1-x2 n) is continuous for all x in

Question

The function f: R arrow R defined by f(x)= displaystyle lim n arrow ∞ ( cos (2 π x)-x2 n sin (x-1)/1+x2 n+1-x2 n) is continuous for all x in (A) R - -1  (B) R - -1,1  (C) R - 1  (D) R- 0

Tardigrade Admin · Accepted Answer

Note: n should be given as a natural number. f(x= begincases (- sin (x-1)/x-1) x< -1 -( sin 2+1) x=-1 cos 2 π x -1< x< 1 1 x=1 (- sin (x-1)/x-1) x >1 endcases f(x) is discontinuous at x=-1 and x=1

Q. The function $f : R \to R$ defined by $f (x) = n \to \infty lim \frac{cos ( 2 π x ) - x ^{2 n} sin ( x - 1 )}{1 + x ^{2 n + 1} - x ^{2 n}}$ is continuous for all $x$ in

$R - {- 1}$

$R - {- 1, 1}$

$R - {1}$

$R - {0}$

Note: $n$ should be given as a natural number.
$f (x = ⎩ ⎨ ⎧ \frac{- s i n ( x - 1 )}{x - 1} - (sin 2 + 1) cos 2 π x 1 \frac{- s i n ( x - 1 )}{x - 1} x < - 1 x = - 1 - 1 < x < 1 x = 1 x > 1$
$f (x)$ is discontinuous at $x = - 1$ and $x = 1$

Q. The function f:R→R defined by f(x)=n→∞lim​1+x2n+1−x2ncos(2πx)−x2nsin(x−1)​ is continuous for all x in

R−{−1}

R−{−1,1}

R−{1}

R−{0}