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Question Number 65779 by mathmax by abdo last updated on 03/Aug/19 | ||
$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{{sin}\left({x}^{\mathrm{2}} \right)−{xtan}\left({x}\right)}{\mathrm{1}−{cos}\left(\mathrm{4}{x}\right)} \\ $$ | ||
Commented by mathmax by abdo last updated on 04/Aug/19 | ||
$$\:\:\:{let}\:{f}\left({x}\right)\:={sin}\left({x}^{\mathrm{2}} \right)−{xtanx}\:\:{and}\:{g}\left({x}\right)=\mathrm{1}−{cos}\left(\mathrm{4}{x}\right) \\ $$$${we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)\:={lim}_{{x}\rightarrow\mathrm{0}} \:\:{g}\left({x}\right)=\mathrm{0} \\ $$$${f}^{'} \left({x}\right)\:=\mathrm{2}{xcos}\left({x}^{\mathrm{2}} \right)−\left({tanx}\:+{x}\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:{f}^{'} \left({x}\right)\:=\mathrm{0} \\ $$$${g}^{'} \left({x}\right)\:=\mathrm{4}{sin}\left(\mathrm{4}{x}\right)\:\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {g}^{'} \left({x}\right)\:=\mathrm{0} \\ $$$${f}^{''} \left({x}\right)\:=\mathrm{2}{cos}\left({x}^{\mathrm{2}} \right)−\mathrm{4}{x}^{\mathrm{2}} {sin}\left({x}^{\mathrm{2}} \right)−\left\{\:\mathrm{1}+{tan}^{\mathrm{2}} {x}\:\:+\mathrm{1}+{tan}^{\mathrm{2}} {x}\:+{x}\left(\mathrm{2}{tanx}\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)\right. \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:{f}^{''} \left({x}\right)\:=\mathrm{0} \\ $$$${g}^{''} \left({x}\right)\:=\mathrm{4}{cos}\left(\mathrm{4}{x}\right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} {g}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)\:=\mathrm{4}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{sin}\left({x}^{\mathrm{2}} \right)−{xtanx}}{\mathrm{1}−{cos}\left(\mathrm{4}{x}\right)}\:=\mathrm{0} \\ $$$$ \\ $$ | ||