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Question Number 115750 by shahria14 last updated on 28/Sep/20

Answered by Dwaipayan Shikari last updated on 28/Sep/20

lim_(x→−1) ((log(1+(x+1)^2 ))/((x+1)^2 ))=1 log(1+(x+1)^2 )=(x+1)^2 (approx)

$$\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{log}\left(\mathrm{1}+\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \right)}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{1} \\ $$$$\mathrm{log}\left(\mathrm{1}+\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \right)=\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{approx}\right) \\ $$

Answered by Ar Brandon last updated on 28/Sep/20

L=lim_(x→−1) ((log(x^2 +2x+2))/((x+1)^2 ))=(0/0) =lim_(x→−1) ((2x+2)/(x^2 +2x+2))∙(1/(2x+2))(l′ho^� pital′s rule) =lim_(x→−1) (1/(x^2 +2x+2))=1

$$\mathscr{L}=\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{log}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{2}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}\centerdot\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{2}}\left(\mathrm{l}'\mathrm{h}\hat {\mathrm{o}pital}'\mathrm{s}\:\mathrm{rule}\right) \\ $$$$\:\:\:\:\:=\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}=\mathrm{1} \\ $$

Answered by mathmax by abdo last updated on 28/Sep/20

let f(x)=((ln(x^2 +2x+2))/((x+1)^2 )) changement x+1 =t give f(x)=f(t−1) =((ln((t−1)^2 +2(t−1)+2))/t^2 ) =((ln(t^2 −2t+1+2t−2+2))/t^2 ) =((ln( 1+t^2 ))/t^2 ) (x→−1⇔t→0) so f(t−1) ∼(t^2 /t^2 )=1 ⇒lim_(t→o) f(t−1) =1 =lim_(x→−1) f(x)

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{2}\right)}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{changement}\:\mathrm{x}+\mathrm{1}\:=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{t}−\mathrm{1}\right)\:=\frac{\mathrm{ln}\left(\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{2}\left(\mathrm{t}−\mathrm{1}\right)+\mathrm{2}\right)}{\mathrm{t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{ln}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{1}+\mathrm{2t}−\mathrm{2}+\mathrm{2}\right)}{\mathrm{t}^{\mathrm{2}} }\:=\frac{\mathrm{ln}\left(\:\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{t}^{\mathrm{2}} }\:\:\:\left(\mathrm{x}\rightarrow−\mathrm{1}\Leftrightarrow\mathrm{t}\rightarrow\mathrm{0}\right)\:\mathrm{so} \\ $$$$\mathrm{f}\left(\mathrm{t}−\mathrm{1}\right)\:\sim\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} }=\mathrm{1}\:\Rightarrow\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{o}} \mathrm{f}\left(\mathrm{t}−\mathrm{1}\right)\:=\mathrm{1}\:=\mathrm{lim}_{\mathrm{x}\rightarrow−\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right) \\ $$

Answered by TANMAY PANACEA last updated on 28/Sep/20

if we do like this way t=(x+1)^2 lim_(t→0) ((log(1+t))/t)=1

$${if}\:{we}\:{do}\:{like}\:{this}\:{way} \\ $$$${t}=\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:\:\:\:\: \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{log}\left(\mathrm{1}+{t}\right)}{{t}}=\mathrm{1} \\ $$

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