Question Number 200840 by Rydel last updated on 24/Nov/23
$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{{x}^{{n}} \mathrm{sin}\:{a}−{a}^{{n}} \mathrm{sin}\:{x}}{{x}−{a}} \\ $$
Answered by MM42 last updated on 24/Nov/23
$${lim}_{{x}\rightarrow{a}} \:\frac{{x}^{{n}} {sina}−{a}^{{n}} {sina}+{a}^{{n}} {sina}−{a}^{{n}} {sinx}}{{x}−{a}} \\ $$$$={lim}_{{x}\rightarrow{a}} \:\frac{\left({x}^{{n}} −{a}^{{n}} \right){sina}−\left({sinx}−{sina}\right){a}^{{n}} }{{x}−{a}} \\ $$$$={lim}_{{x}\rightarrow{a}} \left(\:\frac{\left({x}−{a}\right)\left({x}^{{n}−\mathrm{1}} +{x}^{{n}−\mathrm{1}} {a}+…+{a}^{{n}−\mathrm{1}} \right){sina}}{{x}−{a}}−\frac{\mathrm{2}{sin}\left(\frac{{x}−{a}}{\mathrm{2}}\right)}{{x}−{a}}×{cos}\left(\frac{{x}+{a}}{\mathrm{2}}\right)×{a}^{{n}} \right) \\ $$$$={na}^{{n}−\mathrm{1}} {sina}−{cosa}\:\:\:\checkmark \\ $$$$ \\ $$
Commented by Rydel last updated on 24/Nov/23
$${thank}\:{you}\:{very}\:{much} \\ $$
Answered by BaliramKumar last updated on 24/Nov/23
$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\frac{\frac{\mathrm{d}}{\mathrm{dx}}\left({x}^{{n}} \mathrm{sin}\:{a}−{a}^{{n}} \mathrm{sin}\:{x}\right)}{\frac{\mathrm{d}}{\mathrm{dx}}\left({x}−{a}\right)}\:=\:\frac{\mathrm{n}{x}^{{n}−\mathrm{1}} {sina}−{a}^{{n}} {cosx}}{\mathrm{1}} \\ $$$${na}^{{n}−\mathrm{1}} {sina}−{a}^{{n}} {cosa}\:=\:\frac{{a}^{{n}} \left({nsina}−{acosa}\right)}{{a}} \\ $$$$ \\ $$