Question

Let $X=\{xϵR|x\ne -3\}$. Let $f:X\to R$ be a function defined as
$f\left(x\right)=\{\begin{array}{cc}\frac{x^3-27}{x^2-9}& ifx\ne 3\\ \frac{9}{2}& ifx=3\end{array}$
Then $f$ is

• A. Continuous at all real numbers
• B. Continuous at all points in the domain
• C. Discontinuous at only $x=-3$
• D. Discontinuous at only $x=3$

Answer 1

Answer: (B)

Continuous at all points in the domain

The given function $f:X\to R$ also $X=xϵR|x\ne -3$

As the function is given at every real number except $x=-3$, So the domain of the function is $R,R\ne (-3)$

The function is $f\left(x\right)=\frac{x^3-27}{x^2-9}ifx\ne 3$

$=\frac{9}{2}ifx=3$

The only breaking point of the function is $3$.

${lim}_{x\to 3}\frac{x^3-27}{x^2-9}=\frac{0}{0}$

As $\frac{0}{0}$ is a un-defined form, Hence using L'Hospital rule to find un-defined limits.

${lim}_{x\to 3}\frac{d(x^3-27)}{d(x^2-9)}=li{m}_{x\to 3}\frac{3x^2}{2x}=li{m}_{x\to 3}\frac{3x}{2}=\frac{9}{2}$

Hence $li{m}_{x\to 3}=\frac{9}{2}$, also The value of function at $x=3$ is $\frac{9}{2}$ $($Given$)$

As the limit and the given value of the function at $x=3$ are same hence the function is continuous at $x=3$,

As $x=3$ was the only breaking point, that's why the function is continuous at all points of domain.

Correct answer is $B$.