Question

The function $\underset{n\to \infty }{lim}{\sin }^{2n}\left(\pi x\right)+[x+\frac{1}{2}]$, where [.] denotes the greatest integer function, is:

• A. continuous at $x=1$ but discontinuous at $x=\frac{3}{2}$
• B. continuous at $x=1$ and $x=\frac{3}{2}$
• C. discontinuous at $x=1$ and $x=\frac{3}{2}$
• D. discontinuous at $x=1$ but continuous at $x=\frac{3}{2}$

Answer 1

The correct option is A continuous at $x=1$ but discontinuous at $x=\frac{3}{2}$

Let $f\left(x\right)=\underset{n\to \infty }{lim}{\sin }^{2n}\left(\pi x\right)+[x+\frac{1}{2}]$

$\Rightarrow f\left(x\right)=\underset{n\to \infty }{lim}\left({\sin }^2\right(\pi x){)}^n+[x+\frac{1}{2}]$
Now, $g\left(x\right)=\underset{n\to \infty }{lim}\left({\sin }^2\right(\pi x){)}^n$ is discontinuous when
${\sin }^2\left(\pi x\right)=1$ or $\pi x=(2n+1)\frac{\pi }{2}$ or $x=\frac{(2n+1)}{2},n\in z$
Thus, $g\left(x\right)$ is discontinuous at $x=\frac{3}{2}$.
Also, $h\left(x\right)=[x+\frac{1}{2}]$ is discontinuous at $x=3/2$.
But $f\left(\frac{3}{2}\right)=\underset{n\to \infty }{lim}{\left({\sin }^2\left(\frac{3\pi }{2}\right)\right)}^n+[\frac{3}{2}+\frac{1}{2}]=1+2=3$
$f\left({\frac{3}{2}}^{+}\right)=\underset{n\to \infty }{lim}{\left({\sin }^2\left({\left(\frac{3\pi }{2}\right)}^{+}\right)\right)}^n+[{\left(\frac{3}{2}\right)}^{+}+\frac{1}{2}]=0+2=2$
Hence, $f\left(x\right)$ is discontinuous at $x=\frac{3}{2}$
Both $g\left(x\right)$ and $h\left(x\right)$ are continuous at $x=1$. Hence, $f\left(x\right)$ is continuous at $x=1$.