Question Number 206095 by RoseAli last updated on 06/Apr/24
Answered by Frix last updated on 07/Apr/24
![lim_(x→0) ((x−sin x)/(x−tan x)) =^([l′Ho^ pital]) lim_(x→0) (((d/dx)[x−sin x])/((d/dx)[x−tan x])) = =lim_(x→0) ((1−cos x)/(−tan^2 x)) Let t=tan (x/2); x→0 ⇒ t→0 cos x =((1−t^2 )/(1+t^2 )); tan x =((2t)/(1−t^2 )) lim_(x→0) ((1−cos x)/(−tan^2 x)) =lim_(t→0) (((2t^2 )/(1+t^2 ))/(−((4t^2 )/((1−t^2 )^2 )))) = =lim_(t→0) −(((1−t^2 )^2 )/(2(t^2 +1))) =−(1/2)](https://www.tinkutara.com/question/Q206102.png)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}−\mathrm{tan}\:{x}}\:\:\overset{\left[\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital}\right]} {=}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}}{{dx}}\left[{x}−\mathrm{sin}\:{x}\right]}{\frac{{d}}{{dx}}\left[{x}−\mathrm{tan}\:{x}\right]}\:= \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}}{−\mathrm{tan}^{\mathrm{2}} \:{x}} \\ $$$$\mathrm{Let}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}};\:{x}\rightarrow\mathrm{0}\:\Rightarrow\:{t}\rightarrow\mathrm{0} \\ $$$$\mathrm{cos}\:{x}\:=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} };\:\mathrm{tan}\:{x}\:=\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:{x}}{−\mathrm{tan}^{\mathrm{2}} \:{x}}\:=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{−\frac{\mathrm{4}{t}^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }}\:= \\ $$$$=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:−\frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by MM42 last updated on 07/Apr/24
$${x}−{sinx}\sim\:\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} \:\:\:\&\:\:\:{x}−{tanx}\sim−\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} }{−\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} }\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\:\checkmark \\ $$