Question Number 211873 by Spillover last updated on 22/Sep/24
Answered by IbtisamAdnan last updated on 23/Sep/24
![lim_(x→4) (((cos𝛂)^x −(sinα)^x −cos 2α)/(x−4)) lim_(x→4) (((cos𝛂)^x . ln cosα − (sin α)^x .ln sinα)/1)[L hospital rule] = (cos^4 α)log (cosα) − (sinα)^4 log (sinα) Answer is b](https://www.tinkutara.com/question/Q211875.png)
$$\:\:\:\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{4}} \frac{\left(\boldsymbol{\mathrm{cos}\alpha}\right)^{\mathrm{x}} −\left(\mathrm{sin}\alpha\right)^{\mathrm{x}} −\mathrm{cos}\:\mathrm{2}\alpha}{\mathrm{x}−\mathrm{4}} \\ $$$$\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{4}} \:\frac{\left(\boldsymbol{\mathrm{cos}\alpha}\right)^{\mathrm{x}} .\:\mathrm{ln}\:\mathrm{cos}\alpha\:\:−\:\left(\mathrm{sin}\:\alpha\right)^{\mathrm{x}} .\mathrm{ln}\:\mathrm{sin}\alpha}{\mathrm{1}}\left[\mathrm{L}\:\mathrm{hospital}\:\mathrm{rule}\right] \\ $$$$=\:\left(\mathrm{cos}\:^{\mathrm{4}} \alpha\right)\mathrm{log}\:\left(\mathrm{cos}\alpha\right)\:−\:\left(\mathrm{sin}\alpha\right)^{\mathrm{4}} \:\mathrm{log}\:\left(\mathrm{sin}\alpha\right) \\ $$$$\mathrm{Answer}\:\mathrm{is}\:\mathrm{b} \\ $$
Commented by Spillover last updated on 23/Sep/24
$${correct}.{thanks} \\ $$