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 111092 by bemath last updated on 02/Sep/20

$$\sqrt{\mathrm{bemath}}$$$$\underset{x\to 0}{\lim }\frac{\arctan \mathrm{x}}{\mathrm{arc}\sin \mathrm{x}-\mathrm{x}}$$

Commented by imsahil last updated on 02/Sep/20

How to write cube root?

Commented by bobhans last updated on 02/Sep/20

$$\sqrt[3]{}$$

Commented by Rasheed.Sindhi last updated on 02/Sep/20

Commented by Rasheed.Sindhi last updated on 02/Sep/20

$$@imsahil$$$$Towritecuberootfromtheapp-keyboard.$$$$\left(1\right)click\sqrt{}\left(markedby1\right)$$$$\left(2\right)clickthebuttonmarkedby2$$$$\left(3\right)write‘‘3{}^{″}$$

Answered by bobhans last updated on 02/Sep/20

Commented by bemath last updated on 02/Sep/20

$$\mathrm{it}\mathrm{should}\mathrm{be}$$$$\underset{x\to 0}{\lim }\frac{\mathrm{arc}\tan \mathrm{x}}{\mathrm{arc}\sin \mathrm{x}-\mathrm{x}}=\underset{x\to 0}{\lim }\frac{\left(\frac{1}{1+{\mathrm{x}}^2}\right)}{(\frac{1}{\sqrt{1-{\mathrm{x}}^2}}-1)}$$$$=1\times \underset{x\to 0}{\lim }\left[\frac{\sqrt{1-{\mathrm{x}}^2}}{1-\sqrt{1-{\mathrm{x}}^2}}\right]$$$$=\underset{x\to 0}{\lim }\left[\frac{1.(1+\sqrt{1-{\mathrm{x}}^2})}{{\mathrm{x}}^2}\right]=\mathrm{\infty }$$

Answered by ajfour last updated on 02/Sep/20

$$\underset{x\to 0}{\lim }\frac{\left(\frac{1}{1+x^2}\right)}{(\frac{1}{\sqrt{1-x^2}}-1)}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com