Question Number 197998 by mathlove last updated on 07/Oct/23
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{3}} }=? \\ $$$${with}\:{out}\:{L}'{Hospital}\:{rul} \\ $$
Answered by witcher3 last updated on 07/Oct/23
![x=sh(t),t→0 =lim_(t→0) ((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 ∀t∈[0,1] we have.... (t^3 /6)≤sh(t)−t≤(t^3 /6)+((sh(1))/(24))t^4 ...proof sh(t)−t−(t^3 /6)=g(t) g^((1)) =ch(t)−1−(t^2 /2) g^((2)) (t)=sh(t)−t=∫_0 ^t ch(t)−1dt≥0 g^((1)) (0)=ch(o)−1≥0 g increase g(0)=0⇒g(t)≥0 sh(t)−t−(t^3 /6)−((sh(1))/(24))t^4 g′(t)=ch(t)−1−(t^2 /2)−((sh(1))/6)t^3 g′′(t)=sh(t)−t−((sh(1))/2)t^2 g′′′(t)=ch(t)−sh(1)t−1 g^((4)) =sh(t)−sh(1)≤0,t∈[0,1] g^((3)) (0)=0⇒g^((3)) ≤0 g^((2)) (0)=0,g^((2)) ≤0 g^((1)) (0)=0⇒g′≤0 g(0)=0⇒g(t)≤0,∀t∈[0,1] (t^3 /6)≤sh(t)−t≤(t^3 /6)+sh(1)(t^4 /(24)) ⇔ ∀t>0 (1/6)((t/(sh(t))))^3 ≤((sh(t)−t)/(sh^3 (t)))≤((t/(sh(t))))^3 ((1/6)+((sh(1))/(24))t)...(E) ((sh(t))/t)→1⇔(t/(sh(t)))→1 (E)⇒lim_(t→0) ((sh(t)−t)/(sh^3 (t)))=(1/6)](https://www.tinkutara.com/question/Q198006.png)
$$\mathrm{x}=\mathrm{sh}\left(\mathrm{t}\right),\mathrm{t}\rightarrow\mathrm{0} \\ $$$$=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sh}\left(\mathrm{t}\right)−\mathrm{t}}{\mathrm{sh}^{\mathrm{3}} \left(\mathrm{t}\right)} \\ $$$$\mathrm{sh}\left(\mathrm{t}\right)−\mathrm{t}=\mathrm{f}\left(\mathrm{t}\right) \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{f}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{f}''\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\mathrm{f}'''\left(\mathrm{t}\right)=\mathrm{ch}\left(\mathrm{0}\right)=\mathrm{1},,\mathrm{claime}\:\forall\mathrm{t}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{we}\:\mathrm{have}…. \\ $$$$\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{6}}\leqslant\mathrm{sh}\left(\mathrm{t}\right)−\mathrm{t}\leqslant\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{6}}+\frac{\mathrm{sh}\left(\mathrm{1}\right)}{\mathrm{24}}\mathrm{t}^{\mathrm{4}} …\mathrm{proof} \\ $$$$\mathrm{sh}\left(\mathrm{t}\right)−\mathrm{t}−\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{6}}=\mathrm{g}\left(\mathrm{t}\right) \\ $$$$\mathrm{g}^{\left(\mathrm{1}\right)} =\mathrm{ch}\left(\mathrm{t}\right)−\mathrm{1}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{g}^{\left(\mathrm{2}\right)} \left(\mathrm{t}\right)=\mathrm{sh}\left(\mathrm{t}\right)−\mathrm{t}=\int_{\mathrm{0}} ^{\mathrm{t}} \mathrm{ch}\left(\mathrm{t}\right)−\mathrm{1dt}\geqslant\mathrm{0} \\ $$$$\mathrm{g}^{\left(\mathrm{1}\right)} \left(\mathrm{0}\right)=\mathrm{ch}\left(\mathrm{o}\right)−\mathrm{1}\geqslant\mathrm{0} \\ $$$$\mathrm{g}\:\mathrm{increase}\:\mathrm{g}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow\mathrm{g}\left(\mathrm{t}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{sh}\left(\mathrm{t}\right)−\mathrm{t}−\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{6}}−\frac{\mathrm{sh}\left(\mathrm{1}\right)}{\mathrm{24}}\mathrm{t}^{\mathrm{4}} \\ $$$$\mathrm{g}'\left(\mathrm{t}\right)=\mathrm{ch}\left(\mathrm{t}\right)−\mathrm{1}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{sh}\left(\mathrm{1}\right)}{\mathrm{6}}\mathrm{t}^{\mathrm{3}} \\ $$$$\mathrm{g}''\left(\mathrm{t}\right)=\mathrm{sh}\left(\mathrm{t}\right)−\mathrm{t}−\frac{\mathrm{sh}\left(\mathrm{1}\right)}{\mathrm{2}}\mathrm{t}^{\mathrm{2}} \\ $$$$\mathrm{g}'''\left(\mathrm{t}\right)=\mathrm{ch}\left(\mathrm{t}\right)−\mathrm{sh}\left(\mathrm{1}\right)\mathrm{t}−\mathrm{1} \\ $$$$\mathrm{g}^{\left(\mathrm{4}\right)} =\mathrm{sh}\left(\mathrm{t}\right)−\mathrm{sh}\left(\mathrm{1}\right)\leqslant\mathrm{0},\mathrm{t}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\mathrm{g}^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)=\mathrm{0}\Rightarrow\mathrm{g}^{\left(\mathrm{3}\right)} \leqslant\mathrm{0} \\ $$$$\mathrm{g}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)=\mathrm{0},\mathrm{g}^{\left(\mathrm{2}\right)} \leqslant\mathrm{0} \\ $$$$\mathrm{g}^{\left(\mathrm{1}\right)} \left(\mathrm{0}\right)=\mathrm{0}\Rightarrow\mathrm{g}'\leqslant\mathrm{0} \\ $$$$\mathrm{g}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow\mathrm{g}\left(\mathrm{t}\right)\leqslant\mathrm{0},\forall\mathrm{t}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{6}}\leqslant\mathrm{sh}\left(\mathrm{t}\right)−\mathrm{t}\leqslant\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{6}}+\mathrm{sh}\left(\mathrm{1}\right)\frac{\mathrm{t}^{\mathrm{4}} }{\mathrm{24}} \\ $$$$\Leftrightarrow\:\forall\mathrm{t}>\mathrm{0}\:\:\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\boldsymbol{\mathrm{t}}}{\boldsymbol{\mathrm{sh}}\left(\boldsymbol{\mathrm{t}}\right)}\right)^{\mathrm{3}} \leqslant\frac{\mathrm{sh}\left(\mathrm{t}\right)−\mathrm{t}}{\boldsymbol{\mathrm{sh}}^{\mathrm{3}} \left(\boldsymbol{\mathrm{t}}\right)}\leqslant\left(\frac{\boldsymbol{\mathrm{t}}}{\boldsymbol{\mathrm{sh}}\left(\boldsymbol{\mathrm{t}}\right)}\right)^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{sh}\left(\mathrm{1}\right)}{\mathrm{24}}\mathrm{t}\right)…\left(\mathrm{E}\right) \\ $$$$\frac{\boldsymbol{\mathrm{sh}}\left(\boldsymbol{\mathrm{t}}\right)}{\boldsymbol{\mathrm{t}}}\rightarrow\mathrm{1}\Leftrightarrow\frac{\mathrm{t}}{\mathrm{sh}\left(\mathrm{t}\right)}\rightarrow\mathrm{1} \\ $$$$\left(\mathrm{E}\right)\Rightarrow\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sh}\left(\mathrm{t}\right)−\mathrm{t}}{\mathrm{sh}^{\mathrm{3}} \left(\mathrm{t}\right)}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$ \\ $$
Commented by universe last updated on 07/Oct/23
$${sir} \\ $$$${what}\:{is}\:{sh}\:?? \\ $$
Commented by witcher3 last updated on 07/Oct/23
$$\mathrm{sh}\left(\mathrm{t}\right)=\frac{\mathrm{e}^{\mathrm{t}} −\mathrm{e}^{−\mathrm{t}} }{\mathrm{2}} \\ $$
Commented by Frix last updated on 07/Oct/23
$$\mathrm{sh}\:{x}\:=\mathrm{sinh}\:{x} \\ $$
Commented by universe last updated on 07/Oct/23
$${thanks}\:{sir} \\ $$
Answered by mr W last updated on 07/Oct/23
$$\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)=\mathrm{sinh}^{−\mathrm{1}} \:{x}={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+\frac{\mathrm{3}{x}^{\mathrm{5}} }{\mathrm{40}}−{o}\left({x}^{\mathrm{5}} \right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{3}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}−{x}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}−\frac{\mathrm{3}{x}^{\mathrm{5}} }{\mathrm{40}}+{o}\left({x}^{\mathrm{5}} \right)}{{x}^{\mathrm{3}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{40}}+{o}\left({x}^{\mathrm{2}} \right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by mathlove last updated on 08/Oct/23
$${tanks} \\ $$
Answered by MathematicalUser2357 last updated on 09/Oct/23
$$\mathrm{Without}\:\mathrm{L}'\mathrm{H}\hat {\mathrm{o}pital}'\mathrm{s}\:\mathrm{rule}, \\ $$$$\mathrm{I}\:\mathrm{can}\:\mathrm{use}\:\mathrm{mean}−\mathrm{limit}\:\mathrm{rule} \\ $$$${f}\left({x}\right)=\frac{{x}−\mathrm{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{3}} } \\ $$$$\frac{{f}\left(−\epsilon\right)+{f}\left(\epsilon\right)}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$