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Question Number 101008 by Rio Michael last updated on 29/Jun/20

$$\mathrm{Given}f\left(x\right)=\{\begin{array}{l}\frac{1}{2}{xe}^{\frac{1}{x}},x\ne 0\\ 0,x=0\end{array}$$$$\mathrm{find}$$$$\left(i\right)\mathrm{Thd}\mathrm{domain}\mathrm{of}f$$$$\left(ii\right)\mathrm{check}\mathrm{the}\mathrm{continuity}\mathrm{of}f\mathrm{at}x=0$$$$\left(iii\right)\mathrm{check}\mathrm{its}\mathrm{differentiability}\mathrm{and}\mathrm{its}\mathrm{sign}$$$$\left(i\right)\mathrm{sketch}\mathrm{this}\mathrm{curve}\mathrm{and}\mathrm{find}\underset{x\to -\mathrm{\infty }}{\lim }f\left(x\right)\mathrm{and}\underset{x\to +\mathrm{\infty }}{\lim }f\left(x\right)$$

Answered by maths mind last updated on 29/Jun/20

$$D_f=\mathbb{R}$$$$2)\underset{x\to 0^{-}}{\lim }f\left(x\right)=0$$$$3)\underset{x\to 0^{+}}{\lim }f\left(x\right)=+\mathrm{\infty }fnotcontinue$$$$\left(iii\right)diffrentiabilite\Rightarrow continuity$$$$\iff (noconginuity\Rightarrow nodifferentiabilitie)$$$$\underset{x\to -\mathrm{\infty }}{\lim }f\left(x\right)=-\mathrm{\infty },\underset{x\to +\mathrm{\infty }}{\lim }f\left(x\right)=+\mathrm{\infty }$$

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