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Question Number 183136 by mathlove last updated on 21/Dec/22

$$\underset{x\to 0}{\lim }\frac{sinx-x}{x^3}=?$$

Answered by TheSupreme last updated on 21/Dec/22

$$sin\left(x\right)=x-\frac{x^3}{6}+o\left(x^4\right)$$$$lim\frac{-\frac{x^3}{6}+o\left(x^4\right)}{x^3}=\frac{-1}{6}$$

Answered by CrispyXYZ last updated on 21/Dec/22

$${\mathrm{L}}^{\prime }{\mathrm{Hopital}}^{\prime }\mathrm{s}\mathrm{rule}$$$$=\underset{x\to 0}{\lim }\frac{\cos x-1}{3x^2}$$$$=\frac{1}{3}\underset{x\to 0}{\lim }\frac{-\sin x}{2x}$$$$=-\frac{1}{6}\underset{x\to 0}{\lim }\frac{\sin x}{x}$$$$=-\frac{1}{6}$$

Commented by TheSupreme last updated on 21/Dec/22

$$sin\left(x\right)-xand{x}^3{don}^{\prime }trespectde{l}^{\prime }{Hopital}^{\prime }sconditions$$

Commented by MJS_new last updated on 21/Dec/22

$$\mathrm{why}\mathrm{not}?\mathrm{with}x=0\mathrm{we}\mathrm{have}$$$$\sin x-x=0\wedge x^3=0$$$$\frac{\frac{d^3[\sin x-x]}{{dx}^3}}{\frac{d^3\left[x^3\right]}{{dx}^3}}=\frac{-\cos x}{6}\underset{x=0}{=}-\frac{1}{6}$$

Answered by CrispyXYZ last updated on 21/Dec/22

$$\mathrm{Another}\mathrm{method}\mathrm{without}\mathrm{Taylor}\mathrm{and}{\mathrm{L}}^{\prime }\mathrm{Hopital}$$$$\mathrm{Let}x=3y$$$$I=\underset{x\to 0}{\lim }\frac{\sin x-x}{x^3}$$$$=\underset{y\to 0}{\lim }\frac{\sin 3y-3y}{{\left(3y\right)}^3}$$$$=\underset{y\to 0}{\lim }\frac{3\sin y-{4\sin }^{3}y-3y}{27y^3}$$$$=\underset{y\to 0}{\lim }\frac{3\sin y-3y}{27y^3}-\underset{y\to 0}{\lim }\frac{{4\sin }^{3}y}{27y^3}$$$$=\frac{1}{9}\underset{y\to 0}{\lim }\frac{\sin y-y}{y^3}-\frac{4}{27}\underset{y\to 0}{\lim }{\left(\frac{\sin y}{y}\right)}^3$$$$=\frac{1}{9}I-\frac{4}{27}$$$$\mathrm{Therefore},I=\frac{1}{9}I-\frac{4}{27}$$$$\Rightarrow I=-\frac{1}{6}$$

Commented by MJS_new last updated on 21/Dec/22

$$\mathrm{nice}.\mathrm{but}\mathrm{you}\mathrm{need}\mathrm{to}\mathrm{show}\underset{y\to 0}{\lim }\frac{\sin y}{y}=1$$

Commented by CrispyXYZ last updated on 21/Dec/22

$$\mathrm{We}\mathrm{can}\mathrm{simply}\mathrm{use}\mathrm{equivalent}\mathrm{infinitesimal}:$$$$\sin x\sim x$$$$\mathrm{So}\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$$$$\mathrm{Another}\mathrm{way}\mathrm{is}\mathrm{to}\mathrm{draw}\mathrm{a}\mathrm{circle}{x}^2+y^2=1$$$$\mathrm{construct}\mathrm{an}\mathrm{angle}\theta \left(\mathrm{rad}\right)\mathrm{so}\mathrm{one}\mathrm{of}\mathrm{the}\mathrm{intersect}\mathrm{is}$$$$(\cos \theta ,\sin \theta ).\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{arc}\mathrm{is}\theta$$$$\mathrm{it}\mathrm{is}\mathrm{easy}\mathrm{to}\mathrm{show}\mathrm{that}$$$$\underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1$$

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