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Question Number 142922 by mathlove last updated on 07/Jun/21
lim_(x→1) ((sin(x+1))/(2x−(√(x^2 +3))))=?
$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{sin}\left({x}+\mathrm{1}\right)}{\mathrm{2}{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}=? \\ $$
Answered by Olaf_Thorendsen last updated on 07/Jun/21
((sin(x+1))/(2x−(√(x^2 +3)))) = ((sin(x+1).(2x+(√(x^2 +3))))/(4x^2 −(x^2 +3))) = ((sin(x+1).(2x+(√(x^2 +3))))/(3(x+1)(x−1))) ∼_1 ((2sin(2))/(3(x−1))) lim_(x→1^− ) ((sin(x+1))/(2x−(√(x^2 +3)))) = −∞ lim_(x→1^+ ) ((sin(x+1))/(2x−(√(x^2 +3)))) = +∞
$$\frac{\mathrm{sin}\left({x}+\mathrm{1}\right)}{\mathrm{2}{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}\:=\:\frac{\mathrm{sin}\left({x}+\mathrm{1}\right).\left(\mathrm{2}{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\right)}{\mathrm{4}{x}^{\mathrm{2}} −\left({x}^{\mathrm{2}} +\mathrm{3}\right)} \\ $$$$=\:\frac{\mathrm{sin}\left({x}+\mathrm{1}\right).\left(\mathrm{2}{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\right)}{\mathrm{3}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)} \\ $$$$\underset{\mathrm{1}} {\sim}\:\frac{\mathrm{2sin}\left(\mathrm{2}\right)}{\mathrm{3}\left({x}−\mathrm{1}\right)} \\ $$$$\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\frac{\mathrm{sin}\left({x}+\mathrm{1}\right)}{\mathrm{2}{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}\:=\:−\infty \\ $$$$\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:\frac{\mathrm{sin}\left({x}+\mathrm{1}\right)}{\mathrm{2}{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}\:=\:+\infty \\ $$
Answered by mathmax by abdo last updated on 07/Jun/21
f(x)=((sin(x+1))/(2x−(√(x^2 +3)))) changement t=x−1 give f(x)=f(t+1)=((sin(t+2))/(2(t+1)−(√((t+1)^2 +3)))) (t→0) =((sin(t+2))/(2t+2−(√(t^2 +2t+4)))) =((sin(t+2))/(2t+2−2(√(1+((t^2 +2t)/4)))))∼((sin(t+2))/(2t+2−2(1+(1/8)(t^2 +2t)))) =((sin(t+2))/(2t+2−2−(1/4)t^2 −(1/2)t))=((sin(t+2))/((3/2)t−(1/4)t^2 )) ⇒lim_(t→0) f(t+1)=∞=lim_(x→1) f(x)
$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{sin}\left(\mathrm{x}+\mathrm{1}\right)}{\mathrm{2x}−\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{3}}}\:\:\mathrm{changement}\:\mathrm{t}=\mathrm{x}−\mathrm{1}\:\mathrm{give} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{t}+\mathrm{1}\right)=\frac{\mathrm{sin}\left(\mathrm{t}+\mathrm{2}\right)}{\mathrm{2}\left(\mathrm{t}+\mathrm{1}\right)−\sqrt{\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}}}\:\:\left(\mathrm{t}\rightarrow\mathrm{0}\right) \\ $$$$=\frac{\mathrm{sin}\left(\mathrm{t}+\mathrm{2}\right)}{\mathrm{2t}+\mathrm{2}−\sqrt{\mathrm{t}^{\mathrm{2}} +\mathrm{2t}+\mathrm{4}}}\:=\frac{\mathrm{sin}\left(\mathrm{t}+\mathrm{2}\right)}{\mathrm{2t}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}+\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{2t}}{\mathrm{4}}}}\sim\frac{\mathrm{sin}\left(\mathrm{t}+\mathrm{2}\right)}{\mathrm{2t}+\mathrm{2}−\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{2t}\right)\right)} \\ $$$$=\frac{\mathrm{sin}\left(\mathrm{t}+\mathrm{2}\right)}{\mathrm{2t}+\mathrm{2}−\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{t}}=\frac{\mathrm{sin}\left(\mathrm{t}+\mathrm{2}\right)}{\frac{\mathrm{3}}{\mathrm{2}}\mathrm{t}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{t}^{\mathrm{2}} }\:\Rightarrow\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{t}+\mathrm{1}\right)=\infty=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right) \\ $$

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