Question

If the function $f(x)=\{\begin{array}{ccc}\frac{x^3-8}{x^2-4},& if& x\ne 2\\ k,& if& x=2\end{array}$ is continuous at $x=2,$ then the value of $k$ is

• A. $2$
• B. $0$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$f(x)=\{\begin{array}{ccc}\frac{x^3-8}{x^2-4},& if& x\ne 2\\ k,& if& x=2\end{array}$
Since, $f(x)$ is continuous at $x=2,$
then $f(2-h)=f(2+h)=f(2),$
i.e, LHL = RHL = value of function at $x=2,$
Now, $LHL=\underset{h\to 0}{\lim }f(x)=\underset{h\to 0}{\lim }f(2-h)$
$=\underset{h\to 0}{\lim }\frac{{(2-h)}^3-8}{{(2-h)}^2-4}$
$=\underset{h\to 0}{\lim }\frac{3{(2-h)}^2}{2(2-h)}$ (By L Hospital rule)
$=\underset{h\to 0}{\lim }\frac{3}{2}(2-h)$
$=\frac{3}{2}.2=3$ $=f(2)=k$
$\therefore$ $k=3$