Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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)

$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left[\left({x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}{x}^{\mathrm{5}} }{\mathrm{15}}\right)−\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+\frac{{x}^{\mathrm{5}} }{\mathrm{120}}\right)\right]−{x}^{\mathrm{3}} }{{x}^{\mathrm{5}} } \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\frac{{x}^{\mathrm{3}} }{\mathrm{2}}+\frac{\mathrm{15}{x}^{\mathrm{5}} }{\mathrm{120}}\right)−{x}^{\mathrm{3}} }{{x}^{\mathrm{5}} }=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com