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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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