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Question Number 119366 by bobhans last updated on 24/Oct/20

lim_(x→−1) (((√(1+(√(x+5))))−(√3))/(x+1)) =?

$$\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\sqrt{{x}+\mathrm{5}}}−\sqrt{\mathrm{3}}}{{x}+\mathrm{1}}\:=? \\ $$

Answered by bobhans last updated on 24/Oct/20

Answered by mathmax by abdo last updated on 24/Oct/20

L(x)=(((√(1+(√(x+5))))−(√3))/(x+1)) we do the changement x+1=t ⇒ L(x)=L(t−1) =(((√(1+(√(t+4))))−(√3))/t) (x→−1 ⇔t→0) ⇒L(t−1)=(((√(1+(√(t+4))))−(√3))((√(1+(√(t+4))))+(√3)))/(t((√(1+(√(t+4))))+(√3)))) =((1+(√(t+4))−3)/(t((√(1+(√(t+4))+(√3)))))) =((((√(t+4))−2)((√(t+4))+2))/(t((√(1+(√(t+4))))+(√3))((√(t+4))+2))) =(t/(t((√(1+(√(t+4))))+(√3))((√(t+4))+2))) ⇒ lim_(t→0) L(t−1) =lim_(t→0) (1/(((√(1+(√(t+4))))+(√3))((√(t+4))+2))) =(1/(2(√3)(4))) =(1/(8(√3))) =((√3)/(24)) ⇒lim_(t→−1) L(x) =((√3)/(24))

$$\mathrm{L}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{1}+\sqrt{\mathrm{x}+\mathrm{5}}}−\sqrt{\mathrm{3}}}{\mathrm{x}+\mathrm{1}}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}+\mathrm{1}=\mathrm{t}\:\Rightarrow \\ $$$$\mathrm{L}\left(\mathrm{x}\right)=\mathrm{L}\left(\mathrm{t}−\mathrm{1}\right)\:=\frac{\sqrt{\mathrm{1}+\sqrt{\mathrm{t}+\mathrm{4}}}−\sqrt{\mathrm{3}}}{\mathrm{t}}\:\:\:\left(\mathrm{x}\rightarrow−\mathrm{1}\:\Leftrightarrow\mathrm{t}\rightarrow\mathrm{0}\right) \\ $$$$\Rightarrow\mathrm{L}\left(\mathrm{t}−\mathrm{1}\right)=\frac{\left.\sqrt{\mathrm{1}+\sqrt{\mathrm{t}+\mathrm{4}}}−\sqrt{\mathrm{3}}\right)\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{t}+\mathrm{4}}}+\sqrt{\mathrm{3}}\right)}{\mathrm{t}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{t}+\mathrm{4}}}+\sqrt{\mathrm{3}}\right)} \\ $$$$=\frac{\mathrm{1}+\sqrt{\mathrm{t}+\mathrm{4}}−\mathrm{3}}{\mathrm{t}\left(\sqrt{\left.\mathrm{1}+\sqrt{\mathrm{t}+\mathrm{4}}+\sqrt{\mathrm{3}}\right)}\right.}\:=\frac{\left(\sqrt{\mathrm{t}+\mathrm{4}}−\mathrm{2}\right)\left(\sqrt{\mathrm{t}+\mathrm{4}}+\mathrm{2}\right)}{\mathrm{t}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{t}+\mathrm{4}}}+\sqrt{\mathrm{3}}\right)\left(\sqrt{\mathrm{t}+\mathrm{4}}+\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{t}}{\mathrm{t}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{t}+\mathrm{4}}}+\sqrt{\mathrm{3}}\right)\left(\sqrt{\mathrm{t}+\mathrm{4}}+\mathrm{2}\right)}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}} \mathrm{L}\left(\mathrm{t}−\mathrm{1}\right)\:=\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{t}+\mathrm{4}}}+\sqrt{\mathrm{3}}\right)\left(\sqrt{\mathrm{t}+\mathrm{4}}+\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}\left(\mathrm{4}\right)}\:=\frac{\mathrm{1}}{\mathrm{8}\sqrt{\mathrm{3}}}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{24}}\:\Rightarrow\mathrm{lim}_{\mathrm{t}\rightarrow−\mathrm{1}} \mathrm{L}\left(\mathrm{x}\right)\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{24}} \\ $$

Answered by 1549442205PVT last updated on 24/Oct/20

Put x+1=t⇒(x→−1∼t→0) lim _(x→−1) (((√(1+(√(x+5))))−(√3))/(x+1)) =lim_(t→0) (((√(1+(√(t+4)) ))−(√3))/t) =lim_(t→0) (((√(t+4))−2)/(t((√(1+(√(t+4)) ))+(√3))))=lim_(t→0) (1/( (√(1+(√(t+4)) ))+(√3))((√(t+4))+2))) =(1/(2(√3).4))=(1/(8(√3)))

$$\mathrm{Put}\:\mathrm{x}+\mathrm{1}=\mathrm{t}\Rightarrow\left(\mathrm{x}\rightarrow−\mathrm{1}\sim\mathrm{t}\rightarrow\mathrm{0}\right) \\ $$$$\underset{\mathrm{x}\rightarrow−\mathrm{1}} {\:\mathrm{lim}\:}\frac{\sqrt{\mathrm{1}+\sqrt{{x}+\mathrm{5}}}−\sqrt{\mathrm{3}}}{{x}+\mathrm{1}}\:=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt{\mathrm{1}+\sqrt{\mathrm{t}+\mathrm{4}}\:}−\sqrt{\mathrm{3}}}{\mathrm{t}} \\ $$$$=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\sqrt{\mathrm{t}+\mathrm{4}}−\mathrm{2}}{\mathrm{t}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{t}+\mathrm{4}}\:}+\sqrt{\mathrm{3}}\right)}=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\left.\:\sqrt{\mathrm{1}+\sqrt{\mathrm{t}+\mathrm{4}}\:}+\sqrt{\mathrm{3}}\right)\left(\sqrt{\mathrm{t}+\mathrm{4}}+\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}.\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{8}\sqrt{\mathrm{3}}} \\ $$

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