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Question Number 94764 by Hamida last updated on 20/May/20

Answered by prakash jain last updated on 20/May/20

1a (i) lim_(x→3) =1 (top point of semicircle) 1a(ii) lim_(x→−1^− ) =0 1a(iii) lim_(x→0^+ ) =0 (iv) function is not continous at f(0) lim_(x→0^− ) f(x)=0=lim_(x→0^+ ) f(x) However f(0) =2 (v) lim_(x→3^− ) f(3)=1=lim_(x→3^+ ) f(x) To makd function continuous f(3) value should be changed to 1.

$$\mathrm{1a}\:\left(\mathrm{i}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}=\mathrm{1}\:\left(\mathrm{top}\:\mathrm{point}\:\mathrm{of}\:\mathrm{semicircle}\right) \\ $$$$\mathrm{1a}\left(\mathrm{ii}\right) \\ $$$$\underset{{x}\rightarrow−\mathrm{1}^{−} } {\mathrm{lim}}=\mathrm{0}\: \\ $$$$\mathrm{1a}\left(\mathrm{iii}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}=\mathrm{0} \\ $$$$\left(\mathrm{iv}\right) \\ $$$$\mathrm{function}\:\mathrm{is}\:\mathrm{not}\:\mathrm{continous}\:\mathrm{at}\:{f}\left(\mathrm{0}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}{f}\left({x}\right)=\mathrm{0}=\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{f}\left({x}\right) \\ $$$$\mathrm{However}\:{f}\left(\mathrm{0}\right)\:=\mathrm{2} \\ $$$$\left(\mathrm{v}\right)\: \\ $$$$\underset{{x}\rightarrow\mathrm{3}^{−} } {\mathrm{lim}}{f}\left(\mathrm{3}\right)=\mathrm{1}=\underset{{x}\rightarrow\mathrm{3}^{+} } {\mathrm{lim}}{f}\left({x}\right) \\ $$$$\mathrm{To}\:\mathrm{makd}\:\mathrm{function}\:\mathrm{continuous} \\ $$$${f}\left(\mathrm{3}\right)\:\mathrm{value}\:\mathrm{should}\:\mathrm{be}\:\mathrm{changed}\:\mathrm{to}\:\mathrm{1}. \\ $$

Commented by prakash jain last updated on 20/May/20

lim_(x→0) (((√(2x^2 +4))−6)/(x^2 −2x−8)) lim_(x→0) (((√(2x^2 +4))−6)/(x^2 −2x−8))×(((√(2x^2 +4))+6)/((√(2x^2 +4))+6)) lim_(x→0) ((2x^2 +4−36)/((x−4)(x+2)))×(1/((√(2x^2 +4))+6)) lim_(x→0) ((2x^2 −32)/((x−4)(x+2)))×(1/((√(2x^2 +4))+6)) lim_(x→0) ((2(x+4)(x−4))/((x−4)(x+2)))×(1/((√(2x^2 +4))+6)) lim_(x→0) ((2(x+4))/((x+2)))×(1/((√(2x^2 +4))+6)) put x=4 =((2×8)/6)×(1/(12))=(2/9)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{8}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{8}}×\frac{\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}}+\mathrm{6}}{\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}}+\mathrm{6}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}−\mathrm{36}}{\left({x}−\mathrm{4}\right)\left({x}+\mathrm{2}\right)}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}}+\mathrm{6}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{32}}{\left({x}−\mathrm{4}\right)\left({x}+\mathrm{2}\right)}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}}+\mathrm{6}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\left({x}+\mathrm{4}\right)\left({x}−\mathrm{4}\right)}{\left({x}−\mathrm{4}\right)\left({x}+\mathrm{2}\right)}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}}+\mathrm{6}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\left({x}+\mathrm{4}\right)}{\left({x}+\mathrm{2}\right)}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}}+\mathrm{6}} \\ $$$${put}\:{x}=\mathrm{4} \\ $$$$=\frac{\mathrm{2}×\mathrm{8}}{\mathrm{6}}×\frac{\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{2}}{\mathrm{9}} \\ $$

Commented by student work last updated on 21/May/20

ans: (1/2)

$$\:\:\:\mathrm{ans}:\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by mathmax by abdo last updated on 21/May/20

let f(x) =(((√(2x^2 +4))−6)/(x^2 −2x−8)) we have x^2 −2x−8 =x^2 −16 −2x+8 =(x−4)(x+4)−2(x−4) =(x−4)(x+4−2) =(x−4)(x+2) lim_(x→4) f(x) =lim_(x→4) (((2x^2 +4−36)/((√(2x^2 +4))+6))/((x−4)(x+2))) =lim_(x→4) ((2x^2 −32)/((x−4)(x+2)((√(2x^2 +4))+6))) =lim_(x→4) ((2(x−4)(x+4))/((x−4)(x+2)((√(2x^2 +4))+6))) =lim_(x→4) ((2x+8)/((x+2)((√(2x^2 +4))+6))) =((16)/(4(12))) =((4×4)/(4×3×4)) =(1/3)

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\sqrt{\mathrm{2x}^{\mathrm{2}} +\mathrm{4}}−\mathrm{6}}{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{8}}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{8}\:=\mathrm{x}^{\mathrm{2}} −\mathrm{16}\:−\mathrm{2x}+\mathrm{8} \\ $$$$=\left(\mathrm{x}−\mathrm{4}\right)\left(\mathrm{x}+\mathrm{4}\right)−\mathrm{2}\left(\mathrm{x}−\mathrm{4}\right)\:=\left(\mathrm{x}−\mathrm{4}\right)\left(\mathrm{x}+\mathrm{4}−\mathrm{2}\right)\:=\left(\mathrm{x}−\mathrm{4}\right)\left(\mathrm{x}+\mathrm{2}\right) \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{4}} \mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{4}} \:\:\:\:\frac{\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{4}−\mathrm{36}}{\sqrt{\mathrm{2x}^{\mathrm{2}} +\mathrm{4}}+\mathrm{6}}}{\left(\mathrm{x}−\mathrm{4}\right)\left(\mathrm{x}+\mathrm{2}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{4}} \:\:\:\:\frac{\mathrm{2x}^{\mathrm{2}} −\mathrm{32}}{\left(\mathrm{x}−\mathrm{4}\right)\left(\mathrm{x}+\mathrm{2}\right)\left(\sqrt{\mathrm{2x}^{\mathrm{2}} +\mathrm{4}}+\mathrm{6}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{4}} \:\:\:\frac{\mathrm{2}\left(\mathrm{x}−\mathrm{4}\right)\left(\mathrm{x}+\mathrm{4}\right)}{\left(\mathrm{x}−\mathrm{4}\right)\left(\mathrm{x}+\mathrm{2}\right)\left(\sqrt{\mathrm{2x}^{\mathrm{2}} +\mathrm{4}}+\mathrm{6}\right)}\: \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{4}} \:\:\:\frac{\mathrm{2x}+\mathrm{8}}{\left(\mathrm{x}+\mathrm{2}\right)\left(\sqrt{\mathrm{2x}^{\mathrm{2}} +\mathrm{4}}+\mathrm{6}\right)}\:=\frac{\mathrm{16}}{\mathrm{4}\left(\mathrm{12}\right)}\:=\frac{\mathrm{4}×\mathrm{4}}{\mathrm{4}×\mathrm{3}×\mathrm{4}}\:=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by prakash jain last updated on 21/May/20

=lim_(x→4) ((2x+8)/((x+2)((√(2x^2 +4))+6))) =((16)/(4(12))) =((4×4)/(4×3×4)) =(1/3) Calculation mistake shown in red x+2=6 so it should be=((16)/(6×12))=(2/9)

$$=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{4}} \:\:\:\frac{\mathrm{2x}+\mathrm{8}}{\left(\mathrm{x}+\mathrm{2}\right)\left(\sqrt{\mathrm{2x}^{\mathrm{2}} +\mathrm{4}}+\mathrm{6}\right)}\:=\frac{\mathrm{16}}{\mathrm{4}\left(\mathrm{12}\right)}\:=\frac{\mathrm{4}×\mathrm{4}}{\mathrm{4}×\mathrm{3}×\mathrm{4}}\:=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{Calculation}\:\mathrm{mistake}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{red} \\ $$$${x}+\mathrm{2}=\mathrm{6} \\ $$$$\mathrm{so}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}=\frac{\mathrm{16}}{\mathrm{6}×\mathrm{12}}=\frac{\mathrm{2}}{\mathrm{9}} \\ $$

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