Question Number 27722 by NECx last updated on 13/Jan/18
$${find}\:{the}\:{limit}\:{of} \\ $$$$ \\ $$$${f}\left({x}\right)=\begin{cases}{\mathrm{1}+{x}\:\:\:\:\:\:\:{x}<\mathrm{1}}\\{{k}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{0}\:\:\:{c}=\mathrm{0}}\\{\mathrm{1}+{x}\:\:\:\:\:,\:{x}>\mathrm{0}}\end{cases} \\ $$
Answered by NECx last updated on 15/Jan/18
$$\:\:{the}\:{limit}\:{is}\:{said}\:{to}\:{exist}\:{is}\:{the} \\ $$$${LHL}={RHL} \\ $$$$ \\ $$$${LHL}=\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\mathrm{1}+{x}=\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$$${RHL}=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\mathrm{1}+{x}=\mathrm{1}+\mathrm{0}=\mathrm{1} \\ $$$$ \\ $$$$\therefore{Since}\:{the}\:{LHL}\neq{RHL},{then} \\ $$$${the}\:{limit}\:{does}\:{not}\:{exist} \\ $$$$ \\ $$