Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 73466 by Rio Michael last updated on 12/Nov/19

$$pleaseexplainthis$$$$\underset{x\to 0}{Lim}\frac{sinx}{x}=1by{l}^{\prime }hopitalstheorem$$$$$$$$\underset{x\to 0}{Lim}\frac{sinx}{x}=0bySqueeztheorem$$$$istheresomethingwrong?$$

Answered by MJS last updated on 12/Nov/19

$$\underset{x\to 0}{\lim }\frac{\sin x}{x}=1\mathrm{by}\mathrm{Squeeze}\mathrm{theorem}:$$$$\mathrm{draw}\mathrm{a}\mathrm{quarter}\mathrm{circle}\mathrm{within}{x}^{+},y^{+}[\mathrm{or}\mathrm{for}$$$$0⩽\alpha ⩽\frac{\pi }{2}]$$$$\mathrm{you}\mathrm{can}\mathrm{see}\mathrm{that}$$$$\tan \alpha ⩾arc⩾\sin \alpha$$$$arc=\alpha$$$$\tan \alpha ⩾\alpha ⩾\sin \alpha$$$$\frac{\sin \alpha }{\cos \alpha }⩾\alpha ⩾\sin \alpha$$$$\frac{\cos \alpha }{\sin \alpha }⩽\frac{1}{\alpha }⩽\frac{1}{\sin \alpha }$$$$\cos \alpha ⩽\frac{\sin \alpha }{\alpha }⩽1$$$$\underset{\alpha \to 0}{\lim }\cos \alpha =1$$$$\Rightarrow \underset{x\to 0}{\lim }\frac{\sin x}{x}=1$$

Commented by Rio Michael last updated on 13/Nov/19

$$thankssir,iwasthinking$$$$iwilluse$$$$-1⩽sinx⩽1$$$$\Rightarrow -\frac{1}{x}⩽\frac{sinx}{x}⩽\frac{1}{x}$$$$\underset{x\to 0}{lim}-\frac{1}{x}⩽\underset{x\to 0}{lim}\frac{sinx}{x}⩽\underset{x\to 0}{lim}\frac{1}{x}$$$$0⩽\underset{x\to 0}{lim}\frac{sinx}{x}⩽0$$$$\therefore \underset{x\to 0}{lim}\frac{sinx}{x}=0$$

Commented by MJS last updated on 13/Nov/19

$$\mathrm{this}\mathrm{is}\mathrm{not}\mathrm{possible}\mathrm{because}$$$$\underset{x\to 0}{\lim }\pm \frac{1}{x}=\pm \mathrm{\infty }$$

Commented by Rio Michael last updated on 13/Nov/19

$$thankssir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com