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  • At , the function <img alt="\mathrm{ y=e^{-|x|}}" src="https://entrancecorner.oncodecogs.co

At \mathrm{x=0}, the function \mathrm{ y=e^{-|x|}} is
 

Option: 1

continuous
 


Option: 2

continuous and differentiable


Option: 3


differentiable with derivative =1

 


Option: 4

differentiable with derivative =-1


Answers (1)

best_answer

\text {RH limit }=\mathrm{\lim _{h \rightarrow 0} e^{-|0+h|}=\lim _{h \rightarrow 0} e^{-h}=1} \text {. }

\text{LH limit }=\mathrm{\lim _{h \rightarrow 0} e^{-|0-h|}=\lim _{h \rightarrow 0} e^{-h}=1.}

So, the function is continuous at  \mathrm{x=0}.

\begin{aligned} & \text { RH derivative }=\mathrm{\lim _{h \rightarrow 0} \frac{e^{-|0+h|}-e^{-|0|}}{h}=\lim _{h \rightarrow 0} \frac{e^{-h}-1}{h}=-1 . }\\ & \text { LH derivative }=\mathrm{\lim _{h \rightarrow 0} \frac{e^{-|0-h|}-e^{-|0|}}{-h}=\lim _{h \rightarrow 0} \frac{e^{-h}-1}{-h}=1 .} \end{aligned}

So, the function is not differentiable at \mathrm{x=0}.

Posted by

shivangi.shekhar

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