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Question Number 55639 by gunawan last updated on 01/Mar/19

Known a ∈ R and  function f : R→R satiesfied  ∣xf(x)+a∣ < sin^2  (x−a).   For all x ∈ R  value of lim_(x→a)  f(x) ..

$$\mathrm{Known}\:{a}\:\in\:\mathbb{R}\:\mathrm{and} \\ $$ $$\mathrm{function}\:{f}\::\:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{satiesfied} \\ $$ $$\mid{xf}\left({x}\right)+{a}\mid\:<\:\mathrm{sin}^{\mathrm{2}} \:\left({x}−{a}\right).\: \\ $$ $$\mathrm{For}\:\mathrm{all}\:{x}\:\in\:\mathbb{R} \\ $$ $$\mathrm{value}\:\mathrm{of}\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{f}\left({x}\right)\:.. \\ $$

Answered by kaivan.ahmadi last updated on 01/Mar/19

−sin^2 (x−a)<xf(x)+a<sin^2 (x−a)  by sandwich theorem  lim_(x→a) xf(x)+a=0⇒lim_(x→a)  xf(x)=−a⇒  lim_(x→a) x×lim_(x→a) f(x)=−a⇒  lim_(x→a) f(x)=−1

$$−{sin}^{\mathrm{2}} \left({x}−{a}\right)<{xf}\left({x}\right)+{a}<{sin}^{\mathrm{2}} \left({x}−{a}\right) \\ $$ $${by}\:{sandwich}\:{theorem} \\ $$ $${li}\underset{{x}\rightarrow{a}} {{m}xf}\left({x}\right)+{a}=\mathrm{0}\Rightarrow{li}\underset{{x}\rightarrow{a}} {{m}}\:{xf}\left({x}\right)=−{a}\Rightarrow \\ $$ $${li}\underset{{x}\rightarrow{a}} {{m}x}×{li}\underset{{x}\rightarrow{a}} {{m}f}\left({x}\right)=−{a}\Rightarrow \\ $$ $${li}\underset{{x}\rightarrow{a}} {{m}f}\left({x}\right)=−\mathrm{1} \\ $$

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