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lim-x-0-1-x-2-8-x-3-1-3-2-x-2-

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Question Number 147459 by liberty last updated on 21/Jul/21
lim_(x→0) (((√(1+x^2 )) ((8+x^3 ))^(1/3) −2)/x^2 ) =?
$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\sqrt[{\mathrm{3}}]{\mathrm{8}+{x}^{\mathrm{3}} }−\mathrm{2}}{{x}^{\mathrm{2}} }\:=? \\ $$
Answered by mathmax by abdo last updated on 21/Jul/21
let f(x)=(((√(1+x^2 )).(^3 (√(8+x^3 )))−2)/x^2 ) ⇒f(x)=(((√(1+x^2 )).(2(1+(x^3 /8))^(1/3) )−2)/x^2 ) ⇒f(x)∼((2(1+(x^2 /2))(1+(x^3 /(24)))−2)/x^2 ) ⇒ f(x)∼((2(1+(x^3 /(24))+(x^2 /2)+(x^5 /(48)))−2)/x^2 ) ⇒f(x)∼(((x^3 /(12))+x^2 +(x^5 /(24)))/x^2 ) ⇒ f(x)∼1+(x/(12))+(x^3 /(24)) ⇒lim_(x→0) f(x)=1
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }.\left(^{\mathrm{3}} \sqrt{\mathrm{8}+\mathrm{x}^{\mathrm{3}} }\right)−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }.\left(\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{8}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \right)−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\mathrm{1}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{24}}\right)−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{24}}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{48}}\right)−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{12}}+\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{x}^{\mathrm{5}} }{\mathrm{24}}}{\mathrm{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\sim\mathrm{1}+\frac{\mathrm{x}}{\mathrm{12}}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{24}}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{1} \\ $$
Answered by gsk2684 last updated on 21/Jul/21
=lim_(x→0) (((1+x^2 )^(1/2) 2(1+(x^3 /8))^(1/3) −2)/x^2 ) =2lim_(x→0) (((1+(x^2 /2)+(((1/2)((1/2)−1))/(2!))x^4 +...)(1+(1/3) (x^3 /8)+...)−1)/x^2 ) =2lim_(x→0) (((1+(x^2 /2))(1)−1)/x^2 ) =1
$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{2}\left(\mathrm{1}+\frac{{x}^{\mathrm{3}} }{\mathrm{8}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{2}}{{x}^{\mathrm{2}} } \\ $$$$=\mathrm{2}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{4}} +…\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\:\frac{{x}^{\mathrm{3}} }{\mathrm{8}}+…\right)−\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$=\mathrm{2}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\mathrm{1}\right)−\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$=\mathrm{1} \\ $$

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