lim-x-0-x-2-2cos-x-tan-x-x-ax-4-1-a-a-1-12-b-1-2-

Question Number 194185 by mnjuly1970 last updated on 29/Jun/23

$$ \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{lim}_{\:{x}\rightarrow\:\mathrm{0}^{\:−} } \:\left\{\:\frac{\:{x}^{\:\mathrm{2}} \:+\mathrm{2}{cos}\left({x}\right)\:+\:\lfloor−\frac{{tan}\left({x}\right)}{{x}}\:\rfloor}{{ax}^{\:\mathrm{4}} }\:\right\}\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}\:=\:? \\ $$$$\:\:\:\:\:\:\:\:{a}:\:\:\:\frac{\mathrm{1}}{\mathrm{12}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}:\:\:\:\mathrm{12}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}:\:\:−\mathrm{12}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$

Answered by MM42 last updated on 29/Jun/23

lim_(x→0^− ) ((x^2 +2cosx−2)/(ax^4 )) = hop→ lim_(x→0^− ) ((2x−2sinx)/(4ax^3 )) = lim_(x→0^− ) ((2((1/6)x^3 ))/(4ax^3 )) = (1/(12a))=1⇒a=(1/(12)) ✓

$${lim}_{{x}\rightarrow\mathrm{0}^{−} } \:\frac{{x}^{\mathrm{2}} +\mathrm{2}{cosx}−\mathrm{2}}{{ax}^{\mathrm{4}} }\:=\:\: \\ $$$${hop}\rightarrow\:{lim}_{{x}\rightarrow\mathrm{0}^{−} } \:\frac{\mathrm{2}{x}−\mathrm{2}{sinx}}{\mathrm{4}{ax}^{\mathrm{3}} }\:=\: \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{−} } \:\frac{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} \right)}{\mathrm{4}{ax}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{12}{a}}=\mathrm{1}\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{12}}\:\checkmark \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 29/Jun/23