Question

The function $f$ defined by $f(x)=\underset{t\to \infty }{\lim }\left\{\frac{(1+\sin \pi x{)}^t-1}{(1+\sin \pi x{)}^t+1}\right\}$ is

• A. every where continuous
• B. discontinuous at all integer values of $x$
• C. continuous at $x=0$
• D. None of these

Answer 1

Answer: (B)

$f(x)={\lim }_{t\to \infty }\left\{\frac{(1+\sin \pi x{)}^t-1}{(1+\sin \pi x{)}^t+1}\right\}$
$\because \sin \pi x>0$ (in I and II quadrants)
$\therefore 2n\pi <\pi x<(2n+1)\pi$
$\Rightarrow 2n<x<2n+1,n\in I$
and sin $\pi x<0$ (in III and IV quadrrants)
$\therefore (2n+1)<\pi x<(2n+2)\pi$
$\Rightarrow 2n+1<x<2n+2,n\in I$ and
$\sin \pi x=0$ if $x=0,1,2,\dots \dots \dots$
$f(x)=\{\begin{array}{ll}\underset{t\to \infty }{\lim }\frac{1-\frac{1}{(1-\sin \pi x{)}^t}}{1+\frac{1}{(1+\sin \pi x{)}^t}},& 2n<x<2n+1\\ \underset{t\to \infty }{\lim }\frac{(1+\sin \pi x{)}^t+1}{(1+\sin \pi x{)}^t+1},& 2n+1<x<2n+2\\ 0,& x=0,1,2,\dots ..\end{array}$
$=\{\begin{array}{ll}1,& 2n<x<2n+1\\ -1& 2n+1<x<2n+2\\ 0,& x=0,1,2\dots \dots \dots .\end{array}$
If $k\in I,f(k)=0$, but $\underset{x\to k}{\lim }f(x)=1$ or $-1$ according as
$k\in (2n,2n+1)$ or $k\in (2n+1,2n+2)$