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Question Number 154081 by iloveisrael last updated on 14/Sep/21

lim_(x→∞) ((32x^5 −14x^4 +3))^(1/5) −((128x^7 +6x^6 −1))^(1/7) =?

$$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\mathrm{5}}]{\mathrm{32}{x}^{\mathrm{5}} −\mathrm{14}{x}^{\mathrm{4}} +\mathrm{3}}−\sqrt[{\mathrm{7}}]{\mathrm{128}{x}^{\mathrm{7}} +\mathrm{6}{x}^{\mathrm{6}} −\mathrm{1}}\:=? \\ $$

Answered by EDWIN88 last updated on 14/Sep/21

lim_(x→∞) 2x((1−((14)/(32x))+(3/(32x^5 ))))^(1/( 5)) −2x((1+(6/(128x))−(1/(128x^7 ))))^(1/7) let (1/x)=φ ∧φ→0 2× lim_(φ→0) ((((1−(7/(16))φ+(3/(32))φ^5 ))^(1/(5 )) −((1+(3/(64))φ−(1/(128))φ^7 ))^(1/(7 )) )/φ) = 2×lim_(φ→0) (((1−(7/(80))φ+(3/(160))φ^5 )−(1+(3/(448))φ−(1/(7×128))φ^7 ))/φ)= 2× (−((211)/(2240)))=−((211)/(1120))

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}2}{x}\sqrt[{\:\mathrm{5}}]{\mathrm{1}−\frac{\mathrm{14}}{\mathrm{32}{x}}+\frac{\mathrm{3}}{\mathrm{32}{x}^{\mathrm{5}} }}−\mathrm{2}{x}\sqrt[{\mathrm{7}}]{\mathrm{1}+\frac{\mathrm{6}}{\mathrm{128}{x}}−\frac{\mathrm{1}}{\mathrm{128}{x}^{\mathrm{7}} }} \\ $$$${let}\:\frac{\mathrm{1}}{{x}}=\phi\:\wedge\phi\rightarrow\mathrm{0} \\ $$$$\:\mathrm{2}×\:\underset{\phi\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\sqrt[{\mathrm{5}\:}]{\mathrm{1}−\frac{\mathrm{7}}{\mathrm{16}}\phi+\frac{\mathrm{3}}{\mathrm{32}}\phi^{\mathrm{5}} }−\sqrt[{\mathrm{7}\:}]{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{64}}\phi−\frac{\mathrm{1}}{\mathrm{128}}\phi^{\mathrm{7}} }}{\phi}\:= \\ $$$$\:\mathrm{2}×\underset{\phi\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\frac{\mathrm{7}}{\mathrm{80}}\phi+\frac{\mathrm{3}}{\mathrm{160}}\phi^{\mathrm{5}} \right)−\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{448}}\phi−\frac{\mathrm{1}}{\mathrm{7}×\mathrm{128}}\phi^{\mathrm{7}} \right)}{\phi}= \\ $$$$\:\mathrm{2}×\:\left(−\frac{\mathrm{211}}{\mathrm{2240}}\right)=−\frac{\mathrm{211}}{\mathrm{1120}} \\ $$

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