Question

Is the function $f$ defined by
$f\left(x\right)=$ $\{\begin{array}{c}x,ifx\le 1\\ 5,ifx>1\end{array}$
continuous at $x=0$? At $x=1$? At $x=2$?

Answer 1

The given function is
$f\left(x\right)=\{\begin{array}{c}x,\text{if}x\le 1\\ 5,\text{if}x>1\end{array}$
At $x=0$
It is evident that$f$ is defined at $0$ and its value at $0$ is $0$.
Then $\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}x=0$
$\therefore$ $\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$
Therefore, $f$ is continuous at $x=0$
At $x=1,$
$f$ is defined at $1$ and its value at $1$ is $1$.
The left hand limit of $f$ at $x=1$ is,
$\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}x=1$
The right hand limit of $f$ at $x=1$ is,
$\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}\left(5\right)=5$
$\therefore$ $\underset{x\to 1}{lim}f\left(x\right)\ne \underset{x\to 1}{lim}f\left(x\right)$
Therefore, $f$ is not continuous at $x=1$.
At $x=2,$
$f$ is defined at $2$ and its value at $2$ is $5$.
Then, $\underset{x\to 2}{lim}f\left(x\right)=\underset{x\to 2}{lim}\left(5\right)=5$
$\therefore \underset{x\to 2}{lim}f\left(x\right)=f\left(2\right)$
Therefore, $f$ is continuous at $x=2$