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Question Number 131371 by EDWIN88 last updated on 04/Feb/21
lim_(x→0) ((2(tan x−sin x)−x^3 )/x^5 ) =?
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\mathrm{tan}\:{x}−\mathrm{sin}\:{x}\right)−{x}^{\mathrm{3}} }{{x}^{\mathrm{5}} }\:=? \\ $$
Answered by liberty last updated on 04/Feb/21
let x = 2t L=lim_(t→0) ((2(tan 2t−sin 2t)−8t^3 )/(32t^5 )) L= lim_(t→0) ((2(((2tan t)/(1−tan^2 t))−((2tan t)/(1+tan^2 t)))−8t^3 )/(32t^5 )) L= lim_(t→0) ((4tan t (2tan^2 t)−8t^3 (1−tan^4 t))/(32t^5 (1−tan^4 t))) L= lim_(t→0) ((tan^3 t−t^3 +t^3 tan^4 t)/(4t^5 )) L=lim_(t→0) ((tan^3 t−t^3 )/(4t^5 ))=lim_(t→0) ((((tan t−t))/t^3 )×((tan^2 t+t tan t +t^2 )/t^2 )) L= (1/4)
$$\mathrm{let}\:\mathrm{x}\:=\:\mathrm{2t}\: \\ $$$$\mathrm{L}=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\mathrm{tan}\:\mathrm{2t}−\mathrm{sin}\:\mathrm{2t}\right)−\mathrm{8t}^{\mathrm{3}} }{\mathrm{32t}^{\mathrm{5}} } \\ $$$$\mathrm{L}=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\frac{\mathrm{2tan}\:\mathrm{t}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{t}}−\frac{\mathrm{2tan}\:\mathrm{t}}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{t}}\right)−\mathrm{8t}^{\mathrm{3}} }{\mathrm{32t}^{\mathrm{5}} } \\ $$$$\mathrm{L}=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4tan}\:\mathrm{t}\:\left(\mathrm{2tan}\:^{\mathrm{2}} \mathrm{t}\right)−\mathrm{8t}^{\mathrm{3}} \left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{4}} \mathrm{t}\right)}{\mathrm{32t}^{\mathrm{5}} \left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{4}} \mathrm{t}\right)} \\ $$$$\mathrm{L}=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:^{\mathrm{3}} \mathrm{t}−\mathrm{t}^{\mathrm{3}} +\mathrm{t}^{\mathrm{3}} \:\mathrm{tan}\:^{\mathrm{4}} \mathrm{t}}{\mathrm{4t}^{\mathrm{5}} } \\ $$$$\mathrm{L}=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:^{\mathrm{3}} \mathrm{t}−\mathrm{t}^{\mathrm{3}} }{\mathrm{4t}^{\mathrm{5}} }=\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\left(\mathrm{tan}\:\mathrm{t}−\mathrm{t}\right)}{\mathrm{t}^{\mathrm{3}} }×\frac{\mathrm{tan}\:^{\mathrm{2}} \mathrm{t}+\mathrm{t}\:\mathrm{tan}\:\mathrm{t}\:+\mathrm{t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{2}} }\right) \\ $$$$\mathrm{L}=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Answered by bemath last updated on 04/Feb/21
= lim_(x→0) ((2[(x+(x^3 /3)+((2x^5 )/(15)))−(x−(x^3 /6)+(x^5 /(120)))]−x^3 )/x^5 ) = lim_(x→0) ((2((x^3 /2)+((15x^5 )/(120)))−x^3 )/x^5 )=(1/4)
$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left[\left({x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}{x}^{\mathrm{5}} }{\mathrm{15}}\right)−\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+\frac{{x}^{\mathrm{5}} }{\mathrm{120}}\right)\right]−{x}^{\mathrm{3}} }{{x}^{\mathrm{5}} } \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\frac{{x}^{\mathrm{3}} }{\mathrm{2}}+\frac{\mathrm{15}{x}^{\mathrm{5}} }{\mathrm{120}}\right)−{x}^{\mathrm{3}} }{{x}^{\mathrm{5}} }=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

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