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Let f(x)=\left\{\begin{array}{ll} \frac{x}{2 x^2+|x|} & , x \neq 0 \\ 1 & , x=0 \end{array} \text {, then } f(x)\right. \text { is }

Option: 1

Continuous but nondifferentiable at x = 0


Option: 2

Is differentiable at x = 0


Option: 3

Discontinuous at x=0


Option: 4

None of these


Answers (1)

best_answer

\begin{aligned} f(0+0) & =\lim _{h \rightarrow 0} f(-h) \\ & =\lim _{h \rightarrow 0} \frac{h}{2 h^2+h} \\ & =\lim _{h \rightarrow 0} \frac{1}{2 h+1}=1 \\ \text { and } f(0-0)= & \lim _{h \rightarrow 0} f(-h) \\ & =\lim _{h \rightarrow 0} \frac{-h}{2 h^2+|-h|} \\ & =\lim _{h \rightarrow 0} \frac{-h}{2 h^2+h} \\ & =\lim _{h \rightarrow 0} \frac{-1}{2 h+1}=-1 \end{aligned}

as f(0 + 0) f(0 – 0), thus f(x) is discontinuous at x = 0

 

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Gautam harsolia

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