Question Number 55639 by gunawan last updated on 01/Mar/19
$$\mathrm{Known}\:{a}\:\in\:\mathbb{R}\:\mathrm{and} \\ $$$$\mathrm{function}\:{f}\::\:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{satiesfied} \\ $$$$\mid{xf}\left({x}\right)+{a}\mid\:<\:\mathrm{sin}^{\mathrm{2}} \:\left({x}−{a}\right).\: \\ $$$$\mathrm{For}\:\mathrm{all}\:{x}\:\in\:\mathbb{R} \\ $$$$\mathrm{value}\:\mathrm{of}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{f}\left({x}\right)\:.. \\ $$
Answered by kaivan.ahmadi last updated on 01/Mar/19
$$−{sin}^{\mathrm{2}} \left({x}−{a}\right)<{xf}\left({x}\right)+{a}<{sin}^{\mathrm{2}} \left({x}−{a}\right) \\ $$$${by}\:{sandwich}\:{theorem} \\ $$$${li}\underset{{x}\rightarrow{a}} {{m}xf}\left({x}\right)+{a}=\mathrm{0}\Rightarrow{li}\underset{{x}\rightarrow{a}} {{m}}\:{xf}\left({x}\right)=−{a}\Rightarrow \\ $$$${li}\underset{{x}\rightarrow{a}} {{m}x}×{li}\underset{{x}\rightarrow{a}} {{m}f}\left({x}\right)=−{a}\Rightarrow \\ $$$${li}\underset{{x}\rightarrow{a}} {{m}f}\left({x}\right)=−\mathrm{1} \\ $$