Question

Which of the following options is/are correct?

$f\left(x\right)=\{\begin{array}{cc}\frac{x^2-4x-5}{x-5}& \text{if}x>5\\ 2x-4& \text{if}x\le 5\end{array}$

• A. $\underset{x\to 5+}{lim}f\left(x\right)=6$
• B. $\underset{x\to 5-}{lim}f\left(x\right)=6$
• C. $f$ is not continuous at $x=5$
• D. $f$ is a continuous function

Answer 1

Answer: (A), (B), (D)

$\underset{x\to 5-}{lim}f\left(x\right)=6$
$f$ is a continuous function
$\underset{x\to 5+}{lim}f\left(x\right)=6$

$f\left(x\right)=\{\begin{array}{cc}\frac{x^2-4x-5}{x-5}& ifx>5\\ 2x-4& ifx\le 5\end{array}\underset{x\to 5+}{lim}f\left(x\right)=\underset{x\to 5+}{lim}2x-4=2.5-4=6\underset{x\to 5-}{lim}f\left(x\right)=\underset{x\to 5-}{lim}\frac{x^2-4x-5}{x-5}=\underset{x\to 5-}{lim}\frac{2x-4}{1}=\frac{2.5-4}{1}=6$

Hence $f\left(x\right)$ is continuous at $x=5$