lim_(x→0) ((x−ln(x+(√(1+x^2 ))))/x^3 )=? with out L′Hospital rul


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Question Number 197998 by mathlove last updated on 07/Oct/23

$\lim_{x \to 0} \frac{x - l n (x + \sqrt{1 + x^{2}})}{x^{3}} = ?$ $w i t h o u t L^{'} H o s p i t a l r u l$
	Answered by witcher3 last updated on 07/Oct/23

	$x = sh (t), t \to 0$ $= \lim_{t \to 0} \frac{sh (t) - t}{{sh}^{3} (t)}$ $sh (t) - t = f (t)$ $f (0) = 0$ $f^{'} (0) = 0$ $f^{″} (0) = 0$ $f^{‴} (t) = ch (0) = 1,, claime \forall t \in [0, 1]$ $we have . . . .$ $\frac{t^{3}}{6} ⩽ sh (t) - t ⩽ \frac{t^{3}}{6} + \frac{sh (1)}{24} t^{4} . . . proof$ $sh (t) - t - \frac{t^{3}}{6} = g (t)$ $g^{(1)} = ch (t) - 1 - \frac{t^{2}}{2}$ $g^{(2)} (t) = sh (t) - t = \int_{0}^{t} ch (t) - 1 dt ⩾ 0$ $g^{(1)} (0) = ch (o) - 1 ⩾ 0$ $g increase g (0) = 0 \Rightarrow g (t) ⩾ 0$ $sh (t) - t - \frac{t^{3}}{6} - \frac{sh (1)}{24} t^{4}$ $g^{'} (t) = ch (t) - 1 - \frac{t^{2}}{2} - \frac{sh (1)}{6} t^{3}$ $g^{″} (t) = sh (t) - t - \frac{sh (1)}{2} t^{2}$ $g^{‴} (t) = ch (t) - sh (1) t - 1$ $g^{(4)} = sh (t) - sh (1) ⩽ 0, t \in [0, 1]$ $g^{(3)} (0) = 0 \Rightarrow g^{(3)} ⩽ 0$ $g^{(2)} (0) = 0, g^{(2)} ⩽ 0$ $g^{(1)} (0) = 0 \Rightarrow g^{'} ⩽ 0$ $g (0) = 0 \Rightarrow g (t) ⩽ 0, \forall t \in [0, 1]$ $\frac{t^{3}}{6} ⩽ sh (t) - t ⩽ \frac{t^{3}}{6} + sh (1) \frac{t^{4}}{24}$ $\Leftrightarrow \forall t > 0 \frac{1}{6} {(\frac{t}{sh (t)})}^{3} ⩽ \frac{sh (t) - t}{{sh}^{3} (t)} ⩽ {(\frac{t}{sh (t)})}^{3} (\frac{1}{6} + \frac{sh (1)}{24} t) . . . (E)$ $\frac{sh (t)}{t} \to 1 \Leftrightarrow \frac{t}{sh (t)} \to 1$ $(E) \Rightarrow \lim_{t \to 0} \frac{sh (t) - t}{{sh}^{3} (t)} = \frac{1}{6}$
	Commented by universe last updated on 07/Oct/23

	$s i r$ $w h a t i s s h ? ?$
	Commented by witcher3 last updated on 07/Oct/23

	$sh (t) = \frac{e^{t} - e^{- t}}{2}$
	Commented by Frix last updated on 07/Oct/23

	$sh x = \sinh x$
	Commented by universe last updated on 07/Oct/23

	$t h a n k s s i r$
	Answered by mr W last updated on 07/Oct/23

	$\ln (x + \sqrt{1 + x^{2}}) = \sinh^{- 1} x = x - \frac{x^{3}}{6} + \frac{3 x^{5}}{40} - o (x^{5})$ $\lim_{x \to 0} \frac{x - l n (x + \sqrt{1 + x^{2}})}{x^{3}}$ $= \lim_{x \to 0} \frac{x - x + \frac{x^{3}}{6} - \frac{3 x^{5}}{40} + o (x^{5})}{x^{3}}$ $= \lim_{x \to 0} (\frac{1}{6} - \frac{3 x^{2}}{40} + o (x^{2}))$ $= \frac{1}{6}$
	Commented by mathlove last updated on 08/Oct/23

	$t a n k s$
	Answered by MathematicalUser2357 last updated on 09/Oct/23

	$Without L^{'} H {\hat{o p i t a l}}^{'} s rule,$ $I can use mean - limit rule$ $f (x) = \frac{x - \ln (x + \sqrt{1 + x^{2}})}{x^{3}}$ $\frac{f (- ϵ) + f (ϵ)}{2} = \frac{1}{6}$
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