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bemath-lim-x-0-arctan-x-arc-sin-x-x-




Question Number 111092 by bemath last updated on 02/Sep/20
  (√(bemath))  lim_(x→0)  ((arctan x)/(arc sin x−x))
$$\:\:\sqrt{\mathrm{bemath}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{arctan}\:\mathrm{x}}{\mathrm{arc}\:\mathrm{sin}\:\mathrm{x}−\mathrm{x}} \\ $$
Commented by imsahil last updated on 02/Sep/20
How to write cube root?
Commented by bobhans last updated on 02/Sep/20
()^(1/(3  ))
$$\sqrt[{\mathrm{3}\:\:}]{} \\ $$
Commented by Rasheed.Sindhi last updated on 02/Sep/20
Commented by Rasheed.Sindhi last updated on 02/Sep/20
@ imsahil  To write cuberoot from the app-keyboard.  (1)click  (√(  ))  (marked by 1)  (2)click the button marked by 2  (3)write  “ 3 ”
$$@\:{imsahil} \\ $$$${To}\:{write}\:{cuberoot}\:{from}\:{the}\:{app}-{keyboard}. \\ $$$$\left(\mathrm{1}\right){click}\:\:\sqrt{\:\:}\:\:\left({marked}\:{by}\:\mathrm{1}\right) \\ $$$$\left(\mathrm{2}\right){click}\:{the}\:{button}\:{marked}\:{by}\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right){write}\:\:“\:\mathrm{3}\:'' \\ $$
Answered by bobhans last updated on 02/Sep/20
Commented by bemath last updated on 02/Sep/20
it should be   lim_(x→0)  ((arc tan x)/(arc sin x−x)) = lim_(x→0)  ((((1/(1+x^2 ))))/(((1/( (√(1−x^2 ))))−1)))  = 1× lim_(x→0) [ ((√(1−x^2 ))/(1−(√(1−x^2 )))) ]   = lim_(x→0)  [((1.(1+(√(1−x^2 ))))/x^2 ) ]= ∞
$$\mathrm{it}\:\mathrm{should}\:\mathrm{be}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{arc}\:\mathrm{tan}\:\mathrm{x}}{\mathrm{arc}\:\mathrm{sin}\:\mathrm{x}−\mathrm{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)}{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}−\mathrm{1}\right)} \\ $$$$=\:\mathrm{1}×\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\:\frac{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\:\right]\: \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\frac{\mathrm{1}.\left(\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\right)}{\mathrm{x}^{\mathrm{2}} }\:\right]=\:\infty\: \\ $$
Answered by ajfour last updated on 02/Sep/20
lim_(x→0) ((((1/(1+x^2 ))))/(((1/( (√(1−x^2 ))))−1)))
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}−\mathrm{1}\right)} \\ $$

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