Question

If $f\left(x\right)=\frac{1}{x(3x+1)}$ then at $x=0,f\left(x\right)$ is

• A. continuous
• B. discontinuous
• C. not determined
• D. $\underset{x\to 0}{lim}f\left(x\right)=2$

Answer 1

The correct option is B discontinuous

$f\left(x\right)=\frac{1}{x(3x+1)}$

LHL$=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\frac{1}{x(3x+1)}$

as $x\to 0^{-},\frac{1}{x}\to -\infty$

$=\frac{-\infty }{0+1}=-\infty$

RHL$=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\frac{1}{x(3x+1)}$

as $x\to 0^{+},\frac{1}{x}\to +\infty$

$=\frac{+\infty }{0+1}=\infty$

LHL$\ne$RHL $\Rightarrow$Discontinuous