Question Number 227080 by Spillover last updated on 29/Dec/25
Answered by Kassista last updated on 29/Dec/25
$$\left.{a}\right)\:\underset{{x}\rightarrow−\mathrm{6}} {\mathrm{lim}}\:{f}\left({x}\right)\:=\:\underset{{x}\rightarrow−\mathrm{6}} {\mathrm{lim}}\:\mathrm{7}−\mathrm{4}{x}\:=\:\mathrm{7}−\mathrm{4}\left(−\mathrm{6}\right)=\:\mathrm{31} \\ $$$$ \\ $$$$\left.{b}\right)\:\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:{f}\left({x}\right)\:=\:\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\mathrm{7}−\mathrm{4}{x}\:=\:\mathrm{7}−\mathrm{4}\left(\mathrm{1}\right)=\mathrm{3} \\ $$$$\:\:{and}\:\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}{f}\left({x}\right)\:=\:\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:{x}^{\mathrm{2}} +\mathrm{2}{x}\:=\:\mathrm{1}+\mathrm{2}\left(\mathrm{1}\right)=\mathrm{3} \\ $$$${since}\:\underset{{x}\rightarrow\mathrm{1}−} {\mathrm{lim}}\:{f}\left({x}\right)=\mathrm{1}=\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:{f}\left({x}\right)\:\Rightarrow\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{1} \\ $$$$ \\ $$