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Question Number 185438 by mathlove last updated on 21/Jan/23

lim_(x→0) ((a^x sin (ax)−b^x sin (bx))/(c^x sin (cx)−d^x sin dx)))= ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{a}^{{x}} \mathrm{sin}\:\left({ax}\right)−{b}^{{x}} \mathrm{sin}\:\left({bx}\right)}{\left.{c}^{{x}} \mathrm{sin}\:\left({cx}\right)−{d}^{{x}} \mathrm{sin}\:{dx}\right)}=\:? \\ $$

Answered by hmr last updated on 21/Jan/23

L′ Hopital Rule:  lim_(x→0)  ((a^x ln(a)sin(ax)+a^(x+1) cos(ax)−b^x ln(b)sin(bx)−b^(x+1) cos(bx))/(c^x ln(c)sin(cx)+c^(x+1) cos(cx)−d^x ln(d)sin(dx)−d^(x+1) cos(dx)))  = ((a − b)/(c − d))

$${L}'\:{Hopital}\:{Rule}: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{a}^{{x}} {ln}\left({a}\right){sin}\left({ax}\right)+{a}^{{x}+\mathrm{1}} {cos}\left({ax}\right)−{b}^{{x}} {ln}\left({b}\right){sin}\left({bx}\right)−{b}^{{x}+\mathrm{1}} {cos}\left({bx}\right)}{{c}^{{x}} {ln}\left({c}\right){sin}\left({cx}\right)+{c}^{{x}+\mathrm{1}} {cos}\left({cx}\right)−{d}^{{x}} {ln}\left({d}\right){sin}\left({dx}\right)−{d}^{{x}+\mathrm{1}} {cos}\left({dx}\right)} \\ $$$$=\:\frac{{a}\:−\:{b}}{{c}\:−\:{d}} \\ $$$$ \\ $$

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