lim_(x→∞) ((1+x^4 ))^(1/4) − ((1+x^5 ))^(1/5) =?


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Question Number 118247 by bemath last updated on 16/Oct/20

$\lim_{x \to \infty} \sqrt[4]{1 + x^{4}} - \sqrt[5]{1 + x^{5}} = ?$
	Answered by bobhans last updated on 16/Oct/20

	$s o l v e \lim_{x \to \infty} \sqrt[4]{1 + x^{4}} - \sqrt[5]{1 + x^{5}} .$ $s o l u t i o n :$ $l e t x = \frac{1}{z}; z \to 0$ $\lim_{z \to 0} \sqrt[4]{1 + \frac{1}{z^{4}}} - \sqrt[5]{1 + \frac{1}{z^{5}}} =$ $\lim_{z \to 0} \frac{\sqrt[4]{z^{4} + 1} - \sqrt[5]{z^{5} + 1}}{z} =$ $\lim_{z \to 0} \frac{(\frac{z^{4}}{4} + 1) - (\frac{z^{5}}{5} + 1)}{z} = \lim_{z \to 0} (\frac{\frac{z^{4}}{4} - \frac{z^{5}}{5}}{z})$ $\lim_{z \to 0} \frac{z^{3}}{4} - \frac{z^{4}}{5} = 0$
	Commented by bemath last updated on 16/Oct/20

	$g a v e k u d o s m a n$
	Answered by mathmax by abdo last updated on 16/Oct/20
	$let f(x)=(1+x^4 )^(1/4) −(1+x^5 )^(1/5) ⇒ f(x)=x(1+(1/x^4 ))^(1/4) −x(1+(1/x^5 ))^(1/5) ⇒f(x)∼x{ 1+(1/(4x^4 ))−1−(1/(5x^5 ))} ⇒f(x)∼(1/(4x^3 ))−(1/(5x^4 )) ⇒lim_(x→∞) f(x)=0$
	$let f (x) = {(1 + x^{4})}^{\frac{1}{4}} - {(1 + x^{5})}^{\frac{1}{5}} \Rightarrow$ $f (x) = x {(1 + \frac{1}{x^{4}})}^{\frac{1}{4}} - x {(1 + \frac{1}{x^{5}})}^{\frac{1}{5}} \Rightarrow f (x) \sim x {1 + \frac{1}{{4 x}^{4}} - 1 - \frac{1}{{5 x}^{5}}}$ $\Rightarrow f (x) \sim \frac{1}{{4 x}^{3}} - \frac{1}{{5 x}^{4}} \Rightarrow \lim_{x \to \infty} f (x) = 0$
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