Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 98618 by Ar Brandon last updated on 15/Jun/20

Answered by Ar Brandon last updated on 15/Jun/20

Commented by mr W last updated on 15/Jun/20

$$blackonwhiteisnotbettertoread?$$

Commented by Ar Brandon last updated on 15/Jun/20

OK, understood. Just wanted to try something new.

Answered by maths mind last updated on 15/Jun/20

$$\frac{ln(tg\left(\frac{\pi }{4}\right)-2y)}{y}=\frac{ln(tg\left(\frac{\pi }{4}\right)-2y)-ln\left(1\right)}{y-0}=\frac{\partial }{\partial y}\left(ln(tg\left(\frac{\pi }{4}\right)-2y){\mid }_{y=0}\right)$$$$let\frac{1}{x}=y\iff$$$$\underset{y\to 0}{\lim }{e}^{\frac{ln(tg\left(\frac{\pi }{4}\right)-2y)}{y}}....E$$$$\frac{ln(tg\left(\frac{\pi }{4}\right)-2y)}{y}=\frac{ln(tg\left(\frac{\pi }{4}\right)-2y)-ln\left(1\right)}{y-0}\frac{\partial }{\partial y}(ln\left(tg(\frac{\pi }{4}-2y)\right){\mid }_{y=0}$$$$$$$$=\frac{-2(1+{tg}^2(\frac{\pi }{4}-2y))}{tg(\frac{\pi }{4}-2y)}=-4$$$$E=e^{-4}=\frac{1}{e^4}$$$$$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com