Question Number 128135 by help last updated on 04/Jan/21
Commented by liberty last updated on 04/Jan/21
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{3sin}\:\mathrm{5x}}{\mathrm{x}}\right)^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{4x}}{\mathrm{x}^{\mathrm{2}} }\right)} =\:\mathrm{15}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{16x}^{\mathrm{2}} }{\mathrm{2}}\right)}{\mathrm{x}^{\mathrm{2}} }\right)} \\ $$$$\:=\mathrm{15}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{8x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\right)} \:=\:\mathrm{15}^{\mathrm{8}} \:. \\ $$
Answered by Dwaipayan Shikari last updated on 04/Jan/21
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{3}{sin}\mathrm{5}{x}}{{x}}\right)\left(\frac{\mathrm{1}−{cos}\mathrm{4}{x}}{{x}^{\mathrm{2}} }\right) \\ $$$$=\mathrm{15}.\left(\frac{\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}{x}}{{x}^{\mathrm{2}} }\right)=\mathrm{30}.\left(\frac{\mathrm{4}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)=\mathrm{120} \\ $$
Answered by Dwaipayan Shikari last updated on 04/Jan/21
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{3}{sin}\mathrm{5}{x}}{{x}}\right)^{\left(\frac{\mathrm{1}−{cos}\mathrm{4}{x}}{{x}^{\mathrm{2}} }\right)} \\ $$$$\left(\mathrm{15}\right)^{\frac{\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}{x}}{{x}^{\mathrm{2}} }} =\mathrm{15}^{\mathrm{8}} \\ $$