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Question Number 148323 by liberty last updated on 27/Jul/21

$$\underset{x\to 0}{\lim }\frac{5\sin \mathrm{x}-7\sin 2\mathrm{x}+3\sin 3\mathrm{x}}{\tan \mathrm{x}-\mathrm{x}}=?$$

Answered by EDWIN88 last updated on 27/Jul/21

$$\underset{x\to 0}{\lim }\frac{5\cos x-14\cos 2x+9\cos 3x}{\sec {}^{2}x-1}$$$$\underset{x\to 0}{\lim }\frac{5\cos x-14(2\cos {}^{2}x-1)+9(4\cos {}^{3}x-3\cos x)}{(\sec x-1)(\sec x+1)}$$$$=\underset{x\to 0}{\lim }\frac{5\cos x-28\cos {}^{2}x+14+36\cos {}^{3}x-27\cos x}{(\sec x-1)(\sec x+1)}$$$$=\underset{x\to 0}{\lim }\frac{36\cos {}^{3}x-28\cos {}^{2}x-22\cos x+14}{(\sec x-1)(\sec x+1)}$$$$=\underset{x\to 0}{\lim }\frac{(\cos x-1)(36\cos {}^{2}x+8\cos x-14)}{\left(\frac{1-\cos x}{\cos x}\right)(\sec x+1)}$$$$=\underset{x\to 0}{\lim }\frac{\cos x(36\cos {}^{2}x+8\cos x-14)}{\sec x+1}.\underset{x\to 0}{\lim }\frac{\cos x-1}{1-\cos x}$$$$=-\frac{36+8-14}{2}=-15.$$

Answered by lyubita last updated on 27/Jul/21

$$\underset{x\to 0}{\lim }\frac{5\sin x-7\sin 2x+3\sin 3x}{\tan x-x}$$$$=\underset{x\to 0}{\lim }\frac{5\cos x-14\cos 2x+9\cos 3x}{\sec {}^{2}x-1}\left(l^{\prime }hopital\right)$$$$=\underset{x\to 0}{\lim }\frac{5(1-2\sin {}^2\frac{1}{2}x)-14(1-2\sin {}^{2}x)+9(1-2\sin {}^2\frac{3}{2}x)}{\tan {}^{2}x}$$$$=\underset{x\to 0}{\lim }\frac{-10\sin {}^2\frac{1}{2}x+28\sin {}^{2}x-18\sin {}^2\frac{3}{2}x}{\tan {}^{2}x}$$$$=-\frac{5}{2}+28-\frac{81}{2}$$$$=-15$$$$$$

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