Question

State true or false

Following function is continous at the point $x=-\frac{1}{2}$
$f\left(x\right)=\left\{\begin{array}{c}\frac{4x^2-1}{2x+1},x\ne -\frac{1}{2}\\ -2,x=-\frac{1}{2}\end{array}\right\}$

• A. True
• B. False

Answer 1

The correct option is A True

Given

$f\left(x\right)=\{\begin{array}{ccc}\frac{4x^2-1}{2x+1},& & x\ne -\frac{1}{2}\\ -2& & x=-\frac{1}{2}\end{array}$

$\underset{x\to {\frac{1}{2}}^{-}}{lim}f\left(x\right)=\frac{4{(-\frac{1}{2})}^2-1}{2(-\frac{1}{2})+1}$

This takes the form $\frac{0}{0}$

Using L`Hospitals Rule

$f\left(x\right)=\frac{8x}{2}=4x$

$\underset{x\to {\frac{1}{2}}^{-}}{lim}f\left(x\right)=4(-\frac{1}{2})=-2$

Same happens with $x\to -{\frac{1}{2}}^{+}$

$f(-\frac{1}{2})=-2$

$⟹\underset{x\to {\frac{1}{2}}^{-}}{lim}f\left(x\right)=f\left(x\right)=\underset{x\to {\frac{1}{2}}^{+}}{lim}f\left(x\right)$

Hence the function is continous at $x=-\frac{1}{2}$