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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

black on white is not better to read?

$${black}\:{on}\:{white}\:{is}\:{not}\:{better}\:{to}\:{read}? \\ $$

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

((ln(tg((π/4))−2y))/y)=((ln(tg((π/4))−2y)−ln(1))/(y−0))=(∂/∂y)(ln(tg((π/4))−2y)∣_(y=0) )  let (1/x)=y⇔  lim_(y→0)  e^((ln(tg((π/4))−2y))/y) ....E  ((ln(tg((π/4))−2y))/y)=((ln(tg((π/4))−2y)−ln(1))/(y−0))(∂/∂y)(ln(tg((π/4)−2y))∣_(y=0)     =((−2(1+tg^2 ((π/4)−2y)))/(tg((π/4)−2y)))=−4  E=e^(−4) =(1/e^4 )

$$\frac{{ln}\left({tg}\left(\frac{\pi}{\mathrm{4}}\right)−\mathrm{2}{y}\right)}{{y}}=\frac{{ln}\left({tg}\left(\frac{\pi}{\mathrm{4}}\right)−\mathrm{2}{y}\right)−{ln}\left(\mathrm{1}\right)}{{y}−\mathrm{0}}=\frac{\partial}{\partial{y}}\left({ln}\left({tg}\left(\frac{\pi}{\mathrm{4}}\right)−\mathrm{2}{y}\right)\mid_{{y}=\mathrm{0}} \right) \\ $$$${let}\:\frac{\mathrm{1}}{{x}}={y}\Leftrightarrow \\ $$$$\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{e}^{\frac{{ln}\left({tg}\left(\frac{\pi}{\mathrm{4}}\right)−\mathrm{2}{y}\right)}{{y}}} ....{E} \\ $$$$\frac{{ln}\left({tg}\left(\frac{\pi}{\mathrm{4}}\right)−\mathrm{2}{y}\right)}{{y}}=\frac{{ln}\left({tg}\left(\frac{\pi}{\mathrm{4}}\right)−\mathrm{2}{y}\right)−{ln}\left(\mathrm{1}\right)}{{y}−\mathrm{0}}\frac{\partial}{\partial{y}}\left({ln}\left({tg}\left(\frac{\pi}{\mathrm{4}}−\mathrm{2}{y}\right)\right)\mid_{{y}=\mathrm{0}} \right. \\ $$$$ \\ $$$$=\frac{−\mathrm{2}\left(\mathrm{1}+{tg}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\mathrm{2}{y}\right)\right)}{{tg}\left(\frac{\pi}{\mathrm{4}}−\mathrm{2}{y}\right)}=−\mathrm{4} \\ $$$${E}={e}^{−\mathrm{4}} =\frac{\mathrm{1}}{{e}^{\mathrm{4}} } \\ $$$$ \\ $$$$ \\ $$

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