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Question Number 130843 by EDWIN88 last updated on 29/Jan/21

Without L′Ho^� pital lim_(x→1) ((((x^3 +7))^(1/3) − (√(x^2 +3)))/(x−1)) ?

$$\:{Without}\:{L}'{H}\hat {{o}pital}\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{7}}\:−\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}}{{x}−\mathrm{1}}\:?\: \\ $$

Answered by mathmax by abdo last updated on 29/Jan/21

let f(x)=(((x^3 +7)^(1/3) −(x^2 +3)^(1/2) )/(x−1)) we do the changement x−1=t ⇒ f(x)=f(t+1) =((((t+1)^3 +7)^(1/3) −((t+1)^2 +3)^(1/2) )/t) =(((t^3 +3t^2 +3t +8)^(1/3) −(t^2 +2t+4)^(1/2) )/t) =((8^(1/3) (1+(1/8)(t^3 +3t^2 +3t))^(1/3) −4^(1/2) (1+(1/4)(t^2 +2t))^(1/2) )/t)( t→0) ⇒ f(t+1)∼((2{1+(1/(24))(t^3 +3t^2 +3t)}−2(1+(1/8)(t^2 +2t)))/t) ⇒f(t+1)∼(((1/(12))(t^3 +3t^2 +3t)−(1/4)(t^2 +2t))/t) ⇒f(t+1)∼(1/(12))(t^2 +3t+3)−(1/4)(t+2) ⇒ lim_(t→0) f(t+1) =(3/(12))−(2/4)=(1/4)−(1/2)=−(1/4) ⇒ lim_(x→1) f(x)=−(1/4)

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{x}^{\mathrm{3}} +\mathrm{7}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{x}−\mathrm{1}}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}−\mathrm{1}=\mathrm{t}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{t}+\mathrm{1}\right)\:=\frac{\left(\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{3}} \:+\mathrm{7}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\left(\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{3}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{t}} \\ $$$$=\frac{\left(\mathrm{t}^{\mathrm{3}} \:+\mathrm{3t}^{\mathrm{2}} \:+\mathrm{3t}\:+\mathrm{8}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{2t}+\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{t}} \\ $$$$=\frac{\mathrm{8}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{t}^{\mathrm{3}} \:+\mathrm{3t}^{\mathrm{2}} \:+\mathrm{3t}\right)\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{2t}\right)\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{t}}\left(\:\mathrm{t}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{t}+\mathrm{1}\right)\sim\frac{\mathrm{2}\left\{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{24}}\left(\mathrm{t}^{\mathrm{3}} \:+\mathrm{3t}^{\mathrm{2}} \:+\mathrm{3t}\right)\right\}−\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{2t}\right)\right)}{\mathrm{t}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{t}+\mathrm{1}\right)\sim\frac{\frac{\mathrm{1}}{\mathrm{12}}\left(\mathrm{t}^{\mathrm{3}} \:+\mathrm{3t}^{\mathrm{2}} \:+\mathrm{3t}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{2t}\right)}{\mathrm{t}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{t}+\mathrm{1}\right)\sim\frac{\mathrm{1}}{\mathrm{12}}\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{3t}+\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{t}+\mathrm{2}\right)\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{t}+\mathrm{1}\right)\:=\frac{\mathrm{3}}{\mathrm{12}}−\frac{\mathrm{2}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right)=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented by EDWIN88 last updated on 30/Jan/21

yes..thank you

$${yes}..{thank}\:{you} \\ $$

Commented by mathmax by abdo last updated on 30/Jan/21

you are welcome

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

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