Question

If the function $ f(x)=\left\{ \begin{matrix} \frac{{{x}^{3}}-8}{{{x}^{2}}-4}, & if & x\ne 2 \\ k, & if & x=2 \\ \end{matrix} \right. $ 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)=\left\{ \begin{matrix} \frac{{{x}^{3}}-8}{{{x}^{2}}-4}, & if & x\ne 2 \\ k, & if & x=2 \\ \end{matrix} \right. $
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}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,f(2-h) $
$ =\underset{h\to 0}{\mathop{\lim }}\,\,\frac{{{(2-h)}^{3}}-8}{{{(2-h)}^{2}}-4} $
$ =\underset{h\to 0}{\mathop{\lim }}\,\,\frac{3{{(2-h)}^{2}}}{2(2-h)} $ (By L Hospital rule)
$ =\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{3}{2}\,(2-h) $
$ =\frac{3}{2}\,.2=3 $ $ =f(2)=k $
$ \therefore $ $ k=3 $