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Question Number 194250 by horsebrand11 last updated on 01/Jul/23

$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{for}\mathrm{which}\mathrm{the}\mathrm{limit}$$$$\underset{x\to 0}{\lim }\frac{\sin \left(\mathrm{ax}\right)-{\tan }^{-1}\left(\mathrm{x}\right)-\mathrm{x}}{{\mathrm{x}}^3+{\mathrm{x}}^4}\mathrm{is}\mathrm{finite}$$$$\mathrm{and}\mathrm{then}\mathrm{evaluate}\mathrm{the}\mathrm{limit}$$

Answered by qaz last updated on 01/Jul/23

$$\sin \left(ax\right)=ax-\frac{1}{6}{a}^3{x}^3+...,\arctan x=x-\frac{1}{3}{x}^3+...$$$$\underset{x\to 0}{lim}\frac{(a-2)x+(\frac{1}{3}-\frac{1}{6}{a}^3)x^3}{x^3}=-1,a=2$$

Answered by gatocomcirrose last updated on 02/Jul/23

$$\underset{x\to 0}{\lim }\frac{\mathrm{acos}\left(\mathrm{ax}\right)-\frac{1}{1+{\mathrm{x}}^2}-1}{{3\mathrm{x}}^2+{4\mathrm{x}}^3}=\frac{\mathrm{a}-2}{0}\to \pm \mathrm{\infty }$$$$\mathrm{a}=2\Rightarrow \underset{x\to 0}{\lim }\frac{-{\mathrm{a}}^2\mathrm{sen}\left(\mathrm{ax}\right)+\frac{2\mathrm{x}}{{(1+{\mathrm{x}}^2)}^2}}{6\mathrm{x}+{12\mathrm{x}}^2}$$$$=\underset{x\to 0}{\lim }\frac{-{\mathrm{a}}^3\cos \left(\mathrm{ax}\right)+\frac{2{(1+{\mathrm{x}}^2)}^2-{8\mathrm{x}}^2(1+{\mathrm{x}}^2)}{{(1+{\mathrm{x}}^2)}^4}}{6+24\mathrm{x}}$$$$=\frac{-{\mathrm{a}}^3+2}{6}=\frac{-8+2}{6}=-1$$

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