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Question Number 124802 by john_santu last updated on 06/Dec/20

lim_(x→∞) ((((x+1))^(1/4) − (x)^(1/4) )/( ((x+1))^(1/3) − (x)^(1/3) )) =?

$$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{4}}]{{x}+\mathrm{1}}\:−\:\sqrt[{\mathrm{4}}]{{x}}}{\:\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:−\:\sqrt[{\mathrm{3}}]{{x}}}\:=?\: \\ $$

Answered by bramlexs22 last updated on 06/Dec/20

lim_(x→∞) (((x)^(1/4) (((1+(1/x)))^(1/4) −1))/( (x)^(1/3) (((1+(1/x)))^(1/3) −1))) = lim_(x→∞) (1/( (x)^(1/(12)) )) (((((1+(1/x)))^(1/(4 )) −1)/( ((1+(1/x)))^(1/3) −1))) = 0

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{4}}]{{x}}\:\left(\sqrt[{\mathrm{4}}]{\mathrm{1}+\frac{\mathrm{1}}{{x}}}−\mathrm{1}\right)}{\:\sqrt[{\mathrm{3}}]{{x}}\:\left(\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{1}}{{x}}}−\mathrm{1}\right)}\:= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{12}}]{{x}}}\:\left(\frac{\sqrt[{\mathrm{4}\:}]{\mathrm{1}+\frac{\mathrm{1}}{{x}}}\:−\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{1}}{{x}}}−\mathrm{1}}\right)\:=\:\mathrm{0} \\ $$$$\: \\ $$

Answered by Bird last updated on 06/Dec/20

f(x)=(((x+1)^(1/4) −x^(1/4) )/((x+1)^(1/3) −x^(1/3) )) ⇒ f(x)=((x^(1/4) {(1+(1/x))^(1/4) −1})/(x^(1/3) {(1+(1/x))^(1/3) −1})) ∼(1/x^((1/3)−(1/4)) )×((1/(4x))/(1/(3x))) (x→+∞) f(x)∼(3/4)×(1/x^(1/(12)) )→0 (x→+∞) ⇒lim_(x→+∞) f(x)=0

$${f}\left({x}\right)=\frac{\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} −{x}^{\frac{\mathrm{1}}{\mathrm{4}}} }{\left({x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −{x}^{\frac{\mathrm{1}}{\mathrm{3}}} }\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{{x}^{\frac{\mathrm{1}}{\mathrm{4}}} \left\{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{1}\right\}}{{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \left\{\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{1}\right\}} \\ $$$$\sim\frac{\mathrm{1}}{{x}^{\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}} }×\frac{\frac{\mathrm{1}}{\mathrm{4}{x}}}{\frac{\mathrm{1}}{\mathrm{3}{x}}}\:\:\left({x}\rightarrow+\infty\right) \\ $$$${f}\left({x}\right)\sim\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{1}}{{x}^{\frac{\mathrm{1}}{\mathrm{12}}} }\rightarrow\mathrm{0}\:\left({x}\rightarrow+\infty\right) \\ $$$$\Rightarrow{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=\mathrm{0} \\ $$

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