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 192033 by sonukgindia last updated on 06/May/23

Answered by mehdee42 last updated on 06/May/23

$$L=e^{{lim}_{x\to 0}(\frac{sinx}{x}-1)\frac{1}{x^2}}=e^{{lim}_{x\to 0}\left(\frac{sinx-1}{x^3}\right)}=e^{{lim}_{x\to 0}\left(\frac{-\frac{1}{6}{x}^3}{x^3}\right)}=e^{-\frac{1}{6}}$$

Answered by qaz last updated on 06/May/23

$$\sin x=x-\frac{1}{6}{x}^3+...$$$$\Rightarrow L=\underset{x\to 0}{lim}{(1-\frac{1}{6}{x}^2)}^{\frac{1}{x^2}}=e^{-1/6}$$

Commented by mehdee42 last updated on 06/May/23

$$forthesamereasonofbeingvagio\frac{o}{o}or\frac{\mathrm{\infty }}{\mathrm{\infty }}$$

Commented by Subhi last updated on 06/May/23

$$whywe{can}^{\prime }tapplythelimitforthebaseandthelimitforthepower$$$$thatwillgive{1}^{\mathrm{\infty }}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com