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Question Number 107299 by bobhans last updated on 10/Aug/20

$$✠\mathrm{bobhans}✠$$$$\mathrm{find}\mathrm{without}{\mathrm{L}}^{\prime }\mathrm{Hopital}\mathrm{and}\mathrm{series}$$$$\underset{x\to 0}{\lim }\frac{\mathrm{x}-\sin \mathrm{x}}{{\mathrm{x}}^3}?$$

Answered by john santu last updated on 10/Aug/20

$$\divideontimes \mathrm{JS}\divideontimes$$$$\mathrm{okay}.{\mathrm{let}}^{\prime }\mathrm{go}$$$$\mathrm{L}=\underset{x\to 0}{\lim }\frac{\mathrm{x}-\sin \mathrm{x}}{{\mathrm{x}}^3},\mathrm{let}\mathrm{x}=2\mathrm{y}$$$$\mathrm{L}=\underset{2\mathrm{y}\to 0}{\lim }\frac{2\mathrm{y}-\sin 2\mathrm{y}}{{8\mathrm{y}}^3}=\underset{\mathrm{y}\to 0}{\lim }\frac{2\mathrm{y}-2\sin \mathrm{y}\cos \mathrm{y}}{{8\mathrm{y}}^3}$$$$\mathrm{L}=\frac{1}{4}\underset{\mathrm{y}\to 0}{\lim }\frac{\mathrm{y}-\sin \mathrm{y}+\sin \mathrm{y}-\sin \mathrm{y}\cos \mathrm{y}}{{\mathrm{y}}^3}$$$$4\mathrm{L}=\underset{\mathrm{y}\to 0}{\lim }\frac{\mathrm{y}-\sin \mathrm{y}}{{\mathrm{y}}^3}+\underset{\mathrm{y}\to 0}{\lim }\frac{\sin \mathrm{y}(1-\cos \mathrm{y})}{{\mathrm{y}}^3}$$$$4\mathrm{L}=\mathrm{L}+\underset{x\to 0}{\lim }\frac{2\sin {}^2\left(\frac{\mathrm{y}}{2}\right)}{{\mathrm{y}}^2}$$$$3\mathrm{L}=2\times \frac{1}{4}\iff \mathrm{L}=\frac{1}{6}.$$

Commented by bemath last updated on 10/Aug/20

$$waw...great$$

Commented by bobhans last updated on 10/Aug/20

$$\mathrm{thAnk}\mathrm{you}$$

Commented by malwaan last updated on 10/Aug/20

$$great!$$

Answered by $@y@m last updated on 10/Aug/20

$$Letx=3y$$$$\mathbb{L}=\underset{x\to 0}{\lim }\frac{\mathrm{x}-\sin \mathrm{x}}{{\mathrm{x}}^3}$$$$=\underset{y\to 0}{\lim }\frac{3y-\sin 3y}{27y^3}$$$$=\underset{y\to 0}{\lim }\frac{3y-(3\sin y-4\sin {}^{3}y)}{27y^3}$$$$=\underset{y\to 0}{\lim }\frac{3y-3\sin y+4\sin {}^{3}y}{27y^3}$$$$=\underset{y\to 0}{\lim }\frac{3y-3\sin y}{27y^3}+\frac{4}{27}{\left(\underset{y\to 0}{\lim }\frac{\sin y}{y}\right)}^3$$$$=\frac{1}{9}\underset{y\to 0}{\lim }\frac{y-\sin y}{y^3}+\frac{4}{27}\times 1^3$$$$\mathbb{L}=\frac{1}{9}\mathbb{L}+\frac{4}{27}$$$$\frac{8}{9}\mathbb{L}=\frac{4}{27}$$$$\frac{8}{9}\mathbb{L}=\frac{4}{27}\times \frac{9}{8}$$$$\mathbb{L}=\frac{1}{6}$$

Commented by bemath last updated on 10/Aug/20

$$waw...great...too$$

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