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Question-126722




Question Number 126722 by 0731619177 last updated on 23/Dec/20
Answered by mahdipoor last updated on 23/Dec/20
x=5 { ((x!!!−10>0)),((2x−10=0)) :}  li_(x→5^− ) m ((x!!!−10)/(2x−10))=−∞  li_(x→5^+ ) m ((x!!!−10)/(2x−10))=+∞
$${x}=\mathrm{5\begin{cases}{{x}!!!−\mathrm{10}>\mathrm{0}}\\{\mathrm{2}{x}−\mathrm{10}=\mathrm{0}}\end{cases}} \\ $$$${l}\underset{{x}\rightarrow\mathrm{5}^{−} } {{i}m}\:\frac{{x}!!!−\mathrm{10}}{\mathrm{2}{x}−\mathrm{10}}=−\infty \\ $$$${l}\underset{{x}\rightarrow\mathrm{5}^{+} } {{i}m}\:\frac{{x}!!!−\mathrm{10}}{\mathrm{2}{x}−\mathrm{10}}=+\infty \\ $$$$ \\ $$$$ \\ $$
Commented by 0731619177 last updated on 23/Dec/20
thanks sir
$${thanks}\:{sir} \\ $$

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