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 169294 by cortano1 last updated on 28/Apr/22

Commented by qaz last updated on 28/Apr/22

Same as : lim_(x→0) ((sin tan x−tan sin x)/x^7 )=−(1/(14))

$$\mathrm{Same}\:\mathrm{as}\::\:\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{tan}\:\mathrm{x}−\mathrm{tan}\:\mathrm{sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{7}} }=−\frac{\mathrm{1}}{\mathrm{14}} \\ $$

Commented by cortano1 last updated on 28/Apr/22

your answer wrong

$${your}\:{answer}\:{wrong} \\ $$

Answered by bobhans last updated on 28/Apr/22

lim_(x→0) ((sinh (sin x)−sin (sinh x))/x^7 ) = lim_(x→0) (((x−(1/(15))x^5 +(1/(90))x^7 +O(x^9 ))−(x−(1/(15))x^5 −(1/(90))x^7 +O(x^9 )))/x^7 ) = lim_(x→0) (((1/(45))x^7 +O(x^9 ))/x^7 ) = (1/(45))

$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sinh}\:\left(\mathrm{sin}\:{x}\right)−\mathrm{sin}\:\left(\mathrm{sinh}\:{x}\right)}{{x}^{\mathrm{7}} }\: \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left({x}−\frac{\mathrm{1}}{\mathrm{15}}{x}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{90}}{x}^{\mathrm{7}} +{O}\left({x}^{\mathrm{9}} \right)\right)−\left({x}−\frac{\mathrm{1}}{\mathrm{15}}{x}^{\mathrm{5}} −\frac{\mathrm{1}}{\mathrm{90}}{x}^{\mathrm{7}} +{O}\left({x}^{\mathrm{9}} \right)\right)}{{x}^{\mathrm{7}} } \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\mathrm{45}}{x}^{\mathrm{7}} +{O}\left({x}^{\mathrm{9}} \right)}{{x}^{\mathrm{7}} }\:=\:\frac{\mathrm{1}}{\mathrm{45}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com