Question

Let $f\left(x\right)=\left[{\tan }^{2}x\right]$. Then which of the following is/are true about $f\left(x\right)$?
(where $[.]$ denotes the greatest integer function.)

Answer 1

$f\left(x\right)=\left[{\tan }^{2}x\right]$
$f\left(0\right)=\left[{\tan }^{2}0\right]=\left[0\right]=0$
$\underset{x\to 0^{+}}{\text{lim}}f\left(x\right)=\underset{x\to 0^{+}}{\text{lim}}\left[{\tan }^{2}x\right]=\underset{h\to 0}{lim}\left[{\tan }^2\right(0+h\left)\right]$
$=\underset{h\to 0}{lim}\left[{\tan }^{2}h\right]=0$
$\underset{x\to 0^{-}}{\text{lim}}f\left(x\right)=\underset{x\to 0^{-}}{\text{lim}}\left[{\tan }^{2}x\right]=\underset{h\to 0}{\text{lim}}\left[{\tan }^2(0-h)\right]$
$=\underset{h\to 0}{\text{lim}}\left[{\tan }^{2}h\right]=0$
$\therefore f\left(x\right)$ is continuous at $x=0$
Now,
$f\left(\frac{\pi }{4}\right)=\left[{\tan }^2\frac{\pi }{4}\right]=\left[1\right]=1$
$\underset{x\to {\frac{\pi }{4}}^{+}}{\text{lim}}f\left(x\right)=\underset{x\to {\frac{\pi }{4}}^{+}}{\text{lim}}\left[{\tan }^{2}x\right]=\underset{h\to 0}{lim}\left[{\tan }^2(\frac{\pi }{4}+h)\right]$
$=\left[\right(1^{+}{)}^2]=[1^{+}]=1(\because \tan (\frac{\pi }{4}+h)=\frac{1+\tan h}{1-\tan h}>1)$
$\underset{x\to {\frac{\pi }{4}}^{-}}{\text{lim}}f\left(x\right)=\underset{x\to {\frac{\pi }{4}}^{-}}{\text{lim}}\left[{\tan }^{2}x\right]=\underset{h\to 0}{lim}\left[{\tan }^2(\frac{\pi }{4}-h)\right]$
$=\left[\right(1^{-}{)}^2]=\left[1^{-}\right]=0(\because \tan (\frac{\pi }{4}-h)=\frac{1-\tan h}{1+\tan h}<1)$
$\Rightarrow$ limit does not exist at $x=\frac{\pi }{4}$
$\therefore f\left(x\right)$ is discontinuous at $x=\frac{\pi }{4}$