Question

Examine the continuity of the following function at the point $x=-\frac{1}{2}$.
$f\left(x\right)=\{\begin{array}{c}\frac{4x^2-1}{2x+1}\\ -2,\end{array}\begin{array}{c}x\ne -\frac{1}{2}\\ x=-\frac{1}{2}\end{array}$.

Answer 1

Limit of f as $x\to -\frac{1}{2}$: $\underset{x\to -1/2}{lim}\frac{4x^2-1}{2x+1}=\underset{x\to -1/2}{lim}\frac{(2x+1)(2x-1)}{(2x+1)}$

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

$\Rightarrow$ f is continuous at $x=-\frac{1}{2}$.