Question Number 141685 by mnjuly1970 last updated on 22/May/21
$$\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:……{nice}\:…\:…\:…\:{calculus}….. \\ $$$$\:\:\mathrm{I}{f}\:\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{{tan}\left({x}\right)}{{x}}\:=\:\mathrm{1}\:,\:{prove}\:{that}: \\ $$$$\:\:\:\:\:\:\:\:{lim}\frac{\mathrm{1}}{{x}}\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{tan}\left({x}\right)}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Answered by iloveisrael last updated on 22/May/21
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}−{x}}{{x}^{\mathrm{2}} \:\mathrm{tan}\:{x}}\:= \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\cancel{{x}}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} \right)−\cancel{{x}}}{{x}^{\mathrm{2}} \left({x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} \right)}\:= \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} }{{x}^{\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} \right)}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by greg_ed last updated on 23/May/21
$$\mathrm{true}! \\ $$