Question Number 35727 by abdo mathsup 649 cc last updated on 22/May/18
$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{{sin}\left({shx}\right)\:−{sh}\left({sinx}\right)}{{x}} \\ $$
Commented by abdo mathsup 649 cc last updated on 23/May/18
$${its}\:{beter}\:{to}\:{use}\:{hospital}\:{theorem}\:{in}\:{this}\:{case} \\ $$$${let}\:{put}\:{f}\left({x}\right)={sin}\left({shx}\right)\:−{sh}\left({sinx}\right)\:{and} \\ $$$${g}\left({x}\right)={x}\:{we}\:{have} \\ $$$${f}^{'} \left({x}\right)\:=\:{ch}\left({x}\right){cos}\left({shx}\right)\:−{cos}\left({x}\right)\:{ch}\left({sinx}\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:{f}\left({x}\right)\:=\:\mathrm{1}\:−\mathrm{1}\:=\mathrm{0} \\ $$$${g}^{'} \left({x}\right)\:=\mathrm{1}\:\:{so}\:\:{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{sin}\left({shx}\right)\:−{sh}\left({sinx}\right)}{{x}}\:=\mathrm{0} \\ $$