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  4. The function f defined by f(x)= displaystyle lim t arrow ∞ ((1+ sin π x)t-1/(1+ sin π x)t+1) is

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Q. The function $f$ defined by $f(x)=\displaystyle\lim _{t \rightarrow \infty}\left\{\frac{(1+\sin \pi x)^{t}-1}{(1+\sin \pi x)^{t}+1}\right\}$ is

Continuity and Differentiability

Solution:

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