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Question Number 134741 by zahaku last updated on 06/Mar/21

lim_(→−3) ((x+3)/( (√(∣x^2 +x−6∣))))  ?

$$\underset{\rightarrow−\mathrm{3}} {{lim}}\frac{{x}+\mathrm{3}}{\:\sqrt{\mid{x}^{\mathrm{2}} +{x}−\mathrm{6}\mid}}\:\:? \\ $$

Answered by mathmax by abdo last updated on 07/Mar/21

let f(x)=((x+3)/( (√(∣x^2 +x−6∣)))) changement x+3=t give  f(x)=g(t)=(t/( (√(∣(t−3)^2 +t−3−6∣)))) =(t/( (√(∣t^2 −6t+9+t−9))∣))  (t→0)  =(t/( (√(∣t^2 −5t∣)))) =(t/( (√(∣t∣))(√(∣t−5∣)))) =((t(√(∣t∣)))/(∣t∣(√(∣t−5∣)))) =δ(t)((√(∣t∣))/( (√(∣t−5∣))))  δ(t)=(t/(∣t∣))=+^− 1 ⇒lim_(t→0) g(t)=0=lim_(x→−3)   g(x)

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}+\mathrm{3}}{\:\sqrt{\mid\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{6}\mid}}\:\mathrm{changement}\:\mathrm{x}+\mathrm{3}=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{t}\right)=\frac{\mathrm{t}}{\:\sqrt{\mid\left(\mathrm{t}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{t}−\mathrm{3}−\mathrm{6}\mid}}\:=\frac{\mathrm{t}}{\:\sqrt{\mid\mathrm{t}^{\mathrm{2}} −\mathrm{6t}+\mathrm{9}+\mathrm{t}−\mathrm{9}}\mid}\:\:\left(\mathrm{t}\rightarrow\mathrm{0}\right) \\ $$$$=\frac{\mathrm{t}}{\:\sqrt{\mid\mathrm{t}^{\mathrm{2}} −\mathrm{5t}\mid}}\:=\frac{\mathrm{t}}{\:\sqrt{\mid\mathrm{t}\mid}\sqrt{\mid\mathrm{t}−\mathrm{5}\mid}}\:=\frac{\mathrm{t}\sqrt{\mid\mathrm{t}\mid}}{\mid\mathrm{t}\mid\sqrt{\mid\mathrm{t}−\mathrm{5}\mid}}\:=\delta\left(\mathrm{t}\right)\frac{\sqrt{\mid\mathrm{t}\mid}}{\:\sqrt{\mid\mathrm{t}−\mathrm{5}\mid}} \\ $$$$\delta\left(\mathrm{t}\right)=\frac{\mathrm{t}}{\mid\mathrm{t}\mid}=\overset{−} {+}\mathrm{1}\:\Rightarrow\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}} \mathrm{g}\left(\mathrm{t}\right)=\mathrm{0}=\mathrm{lim}_{\mathrm{x}\rightarrow−\mathrm{3}} \:\:\mathrm{g}\left(\mathrm{x}\right) \\ $$

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