Let $f(x)=\left\{\begin{array}{l}\frac{x}{2 x^2+|x|}, & x \neq 0 \\ 1, & x=0\end{array}\right.$, then f(x) is
Answer & explanation
Correct answer: option 3
$f(0+0)=\lim\limits_{h \rightarrow 0} f(-h)$
$=\lim\limits_{h \rightarrow 0} \frac{h}{2 h^2+h}$
$=\lim\limits_{h \rightarrow 0} \frac{1}{2 h+1}=1$
and $f(0 - 0)=\lim\limits_{h \rightarrow 0} f(-h)$
$=\lim\limits_{h \rightarrow 0} \frac{-h}{2 h^2+|-h|}$
$=\lim\limits_{h \rightarrow 0} \frac{-h}{2 h^2+h}$
$=\lim\limits_{h \rightarrow 0} \frac{-1}{2 h+1}=-1$
as f(0 + 0) ≠ f(0 – 0), thus f(x) is discontinuous at x = 0.