Question

If $f\left(x\right)=\frac{x^2-9}{x^2-2x-3},x\ne 3$ is continuous at $x=3$, then which one of the following is correct?

• A. $f\left(3\right)=0$
• B. $f\left(3\right)=1.5$
• C. $f\left(3\right)=3$
• D. $f\left(3\right)=-1.5$

Answer 1

The correct option is B $f\left(3\right)=1.5$

As it is given that $f\left(x\right)$ is continuous,

$f\left(3\right)={lim}_{x\to 3}\frac{x^2-9}{x^2-2x-3}=\frac{0}{0}$

As this is a $\frac{0}{0}$ form, we apply L'Hospital's rule,

$\therefore f\left(3\right)={lim}_{x\to 3}\frac{2x}{2x-2}=\frac{6}{4}=1.5$