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  • <img alt="\mathrm{\text { Let } f(x)= \begin{cases}x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}, & x \neq 0 . \text { Then } f(x) \text { at } x=0, \text { is } \\ 0, & x=0\end{cases}}" sr

\mathrm{\text { Let } f(x)= \begin{cases}x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}, & x \neq 0 . \text { Then } f(x) \text { at } x=0, \text { is } \\ 0, & x=0\end{cases}}

Option: 1

unbounded 


Option: 2

bounded, but discontinuous 


Option: 3

continuous, but not differentiable 


Option: 4

differentiable. 


Answers (1)

best_answer

\mathrm{\frac{1}{|x|}+\frac{1}{x}=0 \text { if } x<0}

\mathrm{\text { and }=\frac{2}{x} \text { if } x>0}

\mathrm{\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} x e^{-\frac{2}{x}}=0}

\mathrm{\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} x=0 \text { and } f(0)=0}

f (x) is continuous at x = 0

\mathrm{f^{\prime}\left(0^{+}\right)=\lim _{x \rightarrow 0^{+}} e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}=\lim _{x \rightarrow 0^{+}} e^{\frac{-2}{x}}=0}

\mathrm{f^{\prime}\left(0^{-}\right)=\lim _{x \rightarrow 0^{-}} e^0=1 . f^{\prime}(0) \text { does not exist. }}

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jitender.kumar

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