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Question Number 169480 by TOTTI last updated on 01/May/22

Answered by MikeH last updated on 01/May/22

(a) lim_(x→1^− ) g(x) = lim_(x→1) x = 1 (b) lim_(x→1^+ ) g(x) = lim_(x→1) (2−x^2 )=1 ∴ lim_(x→1) g(x) = 1 (c) g(1) = 3 (d) lim_(x→2^− ) g(x) = lim_(x→2) (2−x^2 ) = −2 (e) lim_(x→2^+ ) g(x) = lim_(x→2) (x−3) = −1 hence lim_(x→2) g(x) does not exist.

$$\left(\mathrm{a}\right)\:\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\mathrm{g}\left({x}\right)\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:{x}\:=\:\mathrm{1} \\ $$$$\left(\mathrm{b}\right)\:\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:\mathrm{g}\left({x}\right)\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left(\mathrm{2}−{x}^{\mathrm{2}} \right)=\mathrm{1} \\ $$$$\therefore\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\mathrm{g}\left({x}\right)\:=\:\mathrm{1} \\ $$$$\left(\mathrm{c}\right)\:\mathrm{g}\left(\mathrm{1}\right)\:=\:\mathrm{3} \\ $$$$\left(\mathrm{d}\right)\:\underset{{x}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\:\mathrm{g}\left({x}\right)\:=\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\:=\:−\mathrm{2} \\ $$$$\left(\mathrm{e}\right)\:\underset{{x}\rightarrow\mathrm{2}^{+} } {\mathrm{lim}}\:\mathrm{g}\left({x}\right)\:=\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\left({x}−\mathrm{3}\right)\:=\:−\mathrm{1} \\ $$$$\mathrm{hence}\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\mathrm{g}\left({x}\right)\:\mathrm{does}\:\mathrm{not}\:\mathrm{exist}. \\ $$

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