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 120511 by bramlexs22 last updated on 01/Nov/20

$$\underset{x\to 0}{\lim }\frac{\sqrt{{\left[{\sin }^{-1}\left(2x\right)\right]}^3}}{x\sin \left(\sqrt{x}\right)}?$$

Answered by Olaf last updated on 01/Nov/20

$$\frac{\sqrt{{\arcsin }^3\left(2x\right)}}{x\sin \left(\sqrt{x}\right)}\underset{0^{+}}{\sim }\frac{\sqrt{{\left(2x\right)}^3}}{x\sqrt{x}}=2\sqrt{2}$$

Answered by john santu last updated on 01/Nov/20

$$\underset{x\to 0}{\lim }\frac{\sqrt{{\left[{\sin }^{-1}\left(2x\right)\right]}^3}}{x\sin \left(\sqrt{x}\right)}=\underset{x\to 0}{\lim }\frac{\frac{3}{2}\sqrt{{\sin }^{-1}\left(2x\right)}.\frac{2}{\sqrt{1-4x^2}}}{\sin \sqrt{x}+\frac{x\cos \left(\sqrt{x}\right)}{2\sqrt{x}}}$$$$=\underset{x\to 0}{\lim }\frac{\frac{3}{\sqrt{1-4x^2}}.\sqrt{{\sin }^{-1}\left(2x\right)}}{\sin \sqrt{x}+\frac{1}{2}\sqrt{x}\cos \left(\sqrt{x}\right)}$$$$=6\underset{x\to 0}{\lim }\frac{\sqrt{{\sin }^{-1}\left(2x\right)}}{2\sin \sqrt{x}+\sqrt{x}}=6\underset{x\to 0}{\lim }\frac{\sqrt{2x}}{3\sqrt{x}}=\frac{6\sqrt{2}}{3}=2\sqrt{2}$$$$notethat\{\begin{array}{l}\underset{x\to 0}{\lim }\frac{\sin x}{x}=1\\ \underset{x\to 0}{\lim }\frac{{\sin }^{-1}\left(x\right)}{x}=1\end{array}$$

Answered by bramlexs22 last updated on 01/Nov/20

$$\underset{x\to 0}{\lim }\frac{{\sin }^{-1}\left(2x\right)\sqrt{{\sin }^{-1}\left(2x\right)}}{x\sin \sqrt{x}}=$$$$\underset{x\to 0}{\lim }\frac{{\sin }^{-1}\left(2x\right)}{x}.\underset{x\to 0}{\lim }\frac{\sqrt{\frac{{\sin }^{-1}\left(2x\right)}{x}}}{\frac{\sin \sqrt{x}}{\sqrt{x}}}=$$$$2\times \frac{\sqrt{2}}{1}=2\sqrt{2}$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com