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Question Number 131371 by EDWIN88 last updated on 04/Feb/21

$$\underset{x\to 0}{\lim }\frac{2(\tan x-\sin x)-x^3}{x^5}=?$$

Answered by liberty last updated on 04/Feb/21

$$\mathrm{let}\mathrm{x}=2\mathrm{t}$$$$\mathrm{L}=\underset{\mathrm{t}\to 0}{\lim }\frac{2(\tan 2\mathrm{t}-\sin 2\mathrm{t})-{8\mathrm{t}}^3}{{32\mathrm{t}}^5}$$$$\mathrm{L}=\underset{\mathrm{t}\to 0}{\lim }\frac{2(\frac{2\tan \mathrm{t}}{1-\tan {}^2\mathrm{t}}-\frac{2\tan \mathrm{t}}{1+\tan {}^2\mathrm{t}})-{8\mathrm{t}}^3}{{32\mathrm{t}}^5}$$$$\mathrm{L}=\underset{\mathrm{t}\to 0}{\lim }\frac{4\tan \mathrm{t}\left(2\tan {}^2\mathrm{t}\right)-{8\mathrm{t}}^3(1-\tan {}^4\mathrm{t})}{{32\mathrm{t}}^5(1-\tan {}^4\mathrm{t})}$$$$\mathrm{L}=\underset{\mathrm{t}\to 0}{\lim }\frac{\tan {}^3\mathrm{t}-{\mathrm{t}}^3+{\mathrm{t}}^3\tan {}^4\mathrm{t}}{{4\mathrm{t}}^5}$$$$\mathrm{L}=\underset{\mathrm{t}\to 0}{\lim }\frac{\tan {}^3\mathrm{t}-{\mathrm{t}}^3}{{4\mathrm{t}}^5}=\underset{\mathrm{t}\to 0}{\lim }\left(\frac{(\tan \mathrm{t}-\mathrm{t})}{{\mathrm{t}}^3}\times \frac{\tan {}^2\mathrm{t}+\mathrm{t}\tan \mathrm{t}+{\mathrm{t}}^2}{{\mathrm{t}}^2}\right)$$$$\mathrm{L}=\frac{1}{4}$$

Answered by bemath last updated on 04/Feb/21

$$=\underset{x\to 0}{\lim }\frac{2[(x+\frac{x^3}{3}+\frac{2x^5}{15})-(x-\frac{x^3}{6}+\frac{x^5}{120})]-x^3}{x^5}$$$$=\underset{x\to 0}{\lim }\frac{2(\frac{x^3}{2}+\frac{15x^5}{120})-x^3}{x^5}=\frac{1}{4}$$

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