Without L′Hopital (1)lim_(x→1) ((x(x+(1/x))^5 −32)/(x−1)) =? (2) lim_(x→∞) (√(2x+(√(2x+(√(2x+(√(2x+(√(...)))))))))) −(√(2x)) = ?


Question and Answers Forum

All Questions Topic List
Limits Questions
Previous in All Question Next in All Question
Previous in Limits Next in Limits
Question Number 114981 by bobhans last updated on 22/Sep/20

$W i t h o u t L^{'} H o p i t a l$ $(1) \lim_{x \to 1} \frac{x {(x + \frac{1}{x})}^{5} - 32}{x - 1} = ?$ $(2) \lim_{x \to \infty} \sqrt{2 x + \sqrt{2 x + \sqrt{2 x + \sqrt{2 x + \sqrt{. . .}}}}} - \sqrt{2 x} = ?$
Commented by Dwaipayan Shikari last updated on 22/Sep/20

$\lim_{x \to 1} \frac{x {(x + \frac{1}{x})}^{5} - 2^{5}}{\sqrt[5]{x} (x + \frac{1}{x}) - 2} . \frac{x^{\frac{6}{5}} + x^{- \frac{4}{5}} - 2}{x - 1}$ $\underset{x \to 1}{\lim 5} . 2^{4} . (\frac{2}{5}) = 32 (\frac{x^{\frac{6}{5}} + x^{- \frac{4}{5}} - 2}{x - 1} = \frac{2}{5})$ $\frac{a^{6} + a^{- 4} - 2}{a^{5} - 1} . \frac{a - 1}{a - 1} (x = a^{5})$ $\frac{a^{10} + 1 - 2 a^{4}}{a^{4} + a^{3} + a^{2} + a + 1} . \frac{1}{a^{4} (a - 1)}$ $\frac{1}{5} (a^{10} + 1 - 2 a^{4}) . \frac{1}{a^{5} - a^{4}} = \frac{1}{5} (a^{10} - a^{8} + a^{8} - 2 a^{4} + 1) . \frac{1}{a^{4} (a - 1)} = \frac{1}{5} (a^{5} + a^{4} + \frac{{(a^{4} - 1)}^{2}}{a^{4} (a - 1)}) = \frac{2}{5}$ $[\frac{(a^{4} - 1) (a^{4} - 1)}{a - 1} = \frac{{(a - 1)}^{2} {(a + 1)}^{2} (a^{4} - 1)}{a - 1} = 0]$
Commented by bemath last updated on 23/Sep/20

$(1) c o m p a r e b y H o p i t a l$ $\lim_{x \to 1} \frac{x {(\frac{x^{2} + 1}{x})}^{5} - 32}{x - 1} = \lim_{x \to 1} \frac{{(x^{2} + 1)}^{5} - 32 x^{4}}{x^{4} (x - 1)}$ $\lim_{x \to 1} \frac{5.2 x {(x^{2} + 1)}^{4} - 128 x^{3}}{5 x^{4} - 4 x^{3}} =$ $\frac{10 . 2^{4} - 128}{1} = 160 - 128 = 32$
	Answered by john santu last updated on 22/Sep/20

	$(2) l e t \sqrt{2 x + \sqrt{2 x + \sqrt{2 x + \sqrt{2 x + \sqrt{. . .}}}}} = b$ $\Leftrightarrow 2 x + b = b^{2} \Rightarrow 2 x = b^{2} - b$ $\lim_{b \to \infty} b - \sqrt{b^{2} - b} = \lim_{b \to \infty} \frac{b^{2} - (b^{2} - b)}{b + \sqrt{b^{2} - b}}$ $= \lim_{b \to \infty} \frac{b}{b (1 + \sqrt{1 - \frac{1}{b}})} = \frac{1}{2}$
	Answered by Dwaipayan Shikari last updated on 22/Sep/20

	$\sqrt{2 x + \sqrt{2 x + \sqrt{2 x + . . .}}} = p$ $2 x + p = p^{2}$ $p^{2} - p - 2 x = 0$ $p = \frac{1 \pm \sqrt{1 + 8 x}}{2} = \frac{1 + \sqrt{1 + 8 x}}{2}$ $\frac{1}{2} (1 + \sqrt{1 + 8 x} - 2 \sqrt{2 x}) = \frac{1}{2} + \frac{1 + 8 x - 8 x}{\sqrt{1 + 8 x} + 2 x} = \frac{1}{2}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum