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


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Question Number 163386 by LEKOUMA last updated on 06/Jan/22

$\lim_{x \to - \infty} \sqrt[3]{1 - x^{3}} - \sqrt[4]{x^{4} - 1}$
	Answered by cortano1 last updated on 06/Jan/22
	$Let −x = (1/Y) ; { ((Y→0)),((x=−(1/Y))) :} lim_(Y→0) ((1−(−(1/Y^3 ))))^(1/3) −(((1/Y^4 )−1))^(1/4) = lim_(Y→0) ((((Y^3 +1))^(1/3) −((1−Y^4 ))^(1/4) )/Y) = lim_(Y→0) (((1+(Y^3 /3))−(1−(Y^4 /4)))/Y) = lim_(Y→0) (Y^2 /3) + (Y^3 /4) = 0$
	$L e t - x = \frac{1}{Y}; {\begin{cases} Y \to 0 \\ x = - \frac{1}{Y} \end{cases}$ $\lim_{Y \to 0} \sqrt[3]{1 - (- \frac{1}{Y^{3}})} - \sqrt[4]{\frac{1}{Y^{4}} - 1}$ $= \lim_{Y \to 0} \frac{\sqrt[3]{Y^{3} + 1} - \sqrt[4]{1 - Y^{4}}}{Y}$ $= \lim_{Y \to 0} \frac{(1 + \frac{Y^{3}}{3}) - (1 - \frac{Y^{4}}{4})}{Y}$ $= \lim_{Y \to 0} \frac{Y^{2}}{3} + \frac{Y^{3}}{4} = 0$
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