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Question Number 9960 by Tawakalitu ayo mi last updated on 19/Jan/17

Answered by sandy_suhendra last updated on 20/Jan/17

(i) f(−1)=(−1)^2 +3(−1)−1=−3 lim_(x→−1) x^2 +3x−1=−3 f(x) is continuous at x=−1 because f(−1)=lim_(x→−1) f(x) (ii) f(2)=1 lim_(x→2) ((x^2 −x−2)/(x−2)) = lim_(x→2) (((x−2)(x+1))/((x−2))) =lim_(x→2) (x+1) = 2+1=3 f(x) is discontinuous at x=2 because f(2) ≠ lim_(x→2) f(x)

$$\left(\mathrm{i}\right)\:\mathrm{f}\left(−\mathrm{1}\right)=\left(−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\left(−\mathrm{1}\right)−\mathrm{1}=−\mathrm{3} \\ $$$$\:\:\:\:\:\:\mathrm{li}\underset{\mathrm{x}\rightarrow−\mathrm{1}} {\mathrm{m}}\:\mathrm{x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{1}=−\mathrm{3} \\ $$$$\:\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{at}\:\mathrm{x}=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\mathrm{because}\:\mathrm{f}\left(−\mathrm{1}\right)=\mathrm{li}\underset{\mathrm{x}\rightarrow−\mathrm{1}} {\mathrm{m}}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$ \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{f}\left(\mathrm{2}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{2}} {\mathrm{m}}\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{2}}{\mathrm{x}−\mathrm{2}}\:=\:\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{2}} {\mathrm{m}}\:\frac{\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}+\mathrm{1}\right)}{\left(\mathrm{x}−\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:=\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{2}} {\mathrm{m}}\:\left(\mathrm{x}+\mathrm{1}\right)\:=\:\mathrm{2}+\mathrm{1}=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{discontinuous}\:\mathrm{at}\:\mathrm{x}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\mathrm{because}\:\mathrm{f}\left(\mathrm{2}\right)\:\neq\:\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{2}} {\mathrm{m}}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$

Commented by Tawakalitu ayo mi last updated on 20/Jan/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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