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 170211 by mathlove last updated on 18/May/22

$$\underset{x\to 0}{\lim }\left(\frac{{tan}^{-1}\left(\frac{x-x\sqrt{1-x^2}}{\sqrt{1-x^2}+x^2}\right)}{x^3}\right)=?$$$$pleassolvethis$$

Answered by aleks041103 last updated on 18/May/22

$$1)$$$$\underset{x\to 0}{lim}\frac{arctanx}{x}\stackrel{L^{\prime }H}{=}\underset{x\to 0}{lim}\frac{1}{1+x^2}=1$$$$2)$$$$\underset{x\to 0}{lim}\frac{x-x\sqrt{1-x^2}}{\sqrt{1-x^2}+x^2}=0$$$$3)$$$$\Rightarrow \underset{x\to 0}{\lim }\left(\frac{{tan}^{-1}\left(\frac{x-x\sqrt{1-x^2}}{\sqrt{1-x^2}+x^2}\right)}{x^3}\right)=$$$$=\left[\underset{x\to 0}{lim}\frac{\frac{x-x\sqrt{1-x^2}}{\sqrt{1-x^2}+x^2}}{x^3}\right]\left[\underset{y\to 0}{lim}\frac{arctan\left(y\right)}{y}\right]=$$$$=\underset{x\to 0}{lim}\frac{\frac{x-x\sqrt{1-x^2}}{\sqrt{1-x^2}+x^2}}{x^3}$$$$wherey=\frac{x-x\sqrt{1-x^2}}{\sqrt{1-x^2}+x^2}$$$$4)$$$$\Rightarrow L=\underset{x\to 0}{lim}\frac{x(1-\sqrt{1-x^2})}{x^3(x^2+\sqrt{1-x^2})}=$$$$=\left(\underset{x\to 0}{lim}\frac{1}{x^2+\sqrt{1-x^2}}\right)\left(\underset{x\to 0}{lim}\frac{1-\sqrt{1-x^2}}{x^2}\right)$$$$=\underset{x\to 0}{lim}\frac{1-\sqrt{1-x^2}}{x^2}=\underset{x\to 0}{lim}\frac{(1-\sqrt{1-x^2})(1+\sqrt{1-x^2})}{x^2(1+\sqrt{1-x^2})}=$$$$=\underset{x\to 0}{lim}\frac{1-(1-x^2)}{x^2(1+\sqrt{1-x^2})}=\underset{x\to 0}{lim}\frac{1}{1+\sqrt{1-x^2}}=\frac{1}{2}$$$$\Rightarrow Ans.\frac{1}{2}$$

Commented by mathlove last updated on 18/May/22

$$thanks$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com