Question Number 219806 by SdC355 last updated on 02/May/25
$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left(\frac{\mathrm{cos}\left({x}+{h}\right)}{\mathrm{cos}\left({x}\right)}\right)^{\frac{\mathrm{1}}{{h}}} =?? \\ $$
Answered by fantastic last updated on 02/May/25
$${cos}\left({y}\right)^{\underset{{y}} {\mathrm{1}}} \\ $$
Answered by mehdee7396 last updated on 02/May/25
$${lim}_{{h}\rightarrow\mathrm{0}} \left(\frac{{cos}\left({x}+{h}\right)}{{cosx}}−\mathrm{1}\right)×\frac{\mathrm{1}}{{h}} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}} \left(\frac{{cos}\left({x}+{h}\right)−{cosx}}{{cosx}}\right)×\frac{\mathrm{1}}{{h}} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}} \left(\frac{−\mathrm{2}{sin}\left({x}+\frac{{h}}{\mathrm{2}}\right){sin}\left(\frac{{h}}{\mathrm{2}}\right.}{{cosx}}\right)×\frac{\mathrm{1}}{{h}} \\ $$$$=−{tanx} \\ $$$$\Rightarrow{Ans}={e}^{−{tanx}} \\ $$$$ \\ $$$$ \\ $$