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  • <img alt="\mathrm{f(x)=\left\{\begin{array}{cl} \frac{x}{2 x^2+|x|}, & x \neq 0 \\ 1, & x=0, \text { then } f(x) \text { is } \end{array}\right.}" src="https://entrancecorner.oncodecogs.com

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

Option: 1

continuous but non-differentiable at x=0
 


Option: 2

differentiable at x=0
 


Option: 3

discontinuous at x=0
 


Option: 4

none of these


Answers (1)

best_answer

\mathrm{ \begin{aligned} & f(0+0)=\lim _{h \rightarrow 0} f(h)=\lim _{n \rightarrow 0} \frac{h}{2 h^2+h}=\lim _{n \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) \neq f(0-0), thus f(x) is discontinuous at x=0.

Posted by

Riya

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