Question Number 195905 by cortano12 last updated on 13/Aug/23
$$\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{3}+\mathrm{x}^{\mathrm{4}} }\:−\sqrt{\mathrm{3}+\mathrm{tan}\:^{\mathrm{4}} \mathrm{x}}}{\mathrm{x}^{\mathrm{6}} }\:=? \\ $$
Answered by qaz last updated on 13/Aug/23
$$ \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{x}^{\mathrm{4}} −\mathrm{tan}\:^{\mathrm{4}} {x}}{\mathrm{2}\sqrt{\mathrm{3}}{x}^{\mathrm{6}} }=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left({x}^{\mathrm{2}} +\mathrm{tan}\:^{\mathrm{2}} {x}\right)\left({x}+\mathrm{tan}\:{x}\right)\left({x}−\mathrm{tan}\:{x}\right)}{{x}^{\mathrm{6}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\centerdot\mathrm{2}\centerdot\mathrm{2}\centerdot\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)=−\frac{\mathrm{2}}{\:\mathrm{3}\sqrt{\mathrm{3}}} \\ $$