Question

If $f\left(x\right)=\{\begin{array}{c}x+2,whenx<1\\ 4x-1,when1\le x\le 3\\ x^2+5,whenx>3\end{array}$, then correct statement is-

• A. ${lim}_{x\to 1}f\left(x\right)={lim}_{x\to 3}f\left(x\right)$
• B. $f\left(x\right)$ is continuous at $x=3$
• C. $f\left(x\right)$ is continuous at $x=1$
• D. $f\left(x\right)$ is continuous at $x=1$ and $3$

Answer 1

Answer: (C)

$f\left(x\right)$ is continuous at $x=1$

$f\left(x\right)=x+2$ when $x<1$

$=4x-1$ when $1\le x\le 3$

$=x^2+5$ when $x>3$

We have

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}x+2=3$

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}4x-1=3$

$\underset{x\to 3^{-}}{lim}f\left(x\right)=\underset{x\to 3^{-}}{lim}4x-1=11$

$\underset{x\to 3^{+}}{lim}f\left(x\right)=\underset{x\to 3^{+}}{lim}{x}^2+5=14$

So we can see that $f\left(x\right)$ is continuous at $x=1$ and not at $x=3$