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Question Number 74580 by chess1 last updated on 26/Nov/19

Answered by mind is power last updated on 26/Nov/19

(√(1+t))=1+(t/2)+o(t)⇒(√(x^2 +1))=1+(x^2 /2)+o(x^2 ) (((t+1)))^(1/3) =1+(t/3)+o(t)⇒((1+x^2 ))^(1/3) =1+(x^2 /3)+o(x^2 ) ((1+x^4 ))^(1/4) =1+(x^4 /4)+0(x^4 ) (((1+x^4 )))^(1/5) =1+(x^4 /5)+0(x^4 ) We get(((x^2 /6)+o(x^2 ))/((x^4 /(20))+o(x^4 )))∼((10)/(3x^2 ))→+∞

$$\sqrt{\mathrm{1}+\mathrm{t}}=\mathrm{1}+\frac{\mathrm{t}}{\mathrm{2}}+\mathrm{o}\left(\mathrm{t}\right)\Rightarrow\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}=\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\sqrt[{\mathrm{3}}]{\left(\mathrm{t}+\mathrm{1}\right)}=\mathrm{1}+\frac{\mathrm{t}}{\mathrm{3}}+\mathrm{o}\left(\mathrm{t}\right)\Rightarrow\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }=\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{3}}+\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\sqrt[{\mathrm{4}}]{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }=\mathrm{1}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{4}}+\mathrm{0}\left(\mathrm{x}^{\mathrm{4}} \right) \\ $$$$\sqrt[{\mathrm{5}}]{\left(\mathrm{1}+\mathrm{x}^{\mathrm{4}} \right)}=\mathrm{1}+\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{5}}+\mathrm{0}\left(\mathrm{x}^{\mathrm{4}} \right) \\ $$$$\mathrm{We}\:\mathrm{get}\frac{\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{6}}+\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right)}{\frac{\mathrm{x}^{\mathrm{4}} }{\mathrm{20}}+\mathrm{o}\left(\mathrm{x}^{\mathrm{4}} \right)}\sim\frac{\mathrm{10}}{\mathrm{3x}^{\mathrm{2}} }\rightarrow+\infty \\ $$

Commented by chess1 last updated on 27/Nov/19

thanks

$$\mathrm{thanks} \\ $$

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