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Question Number 159349 by cortano last updated on 15/Nov/21

$$\underset{x\to 0}{\lim }\frac{8\sec x-8+\tan {}^{4}x-4\tan {}^{2}x}{x^6}=?$$

Commented by bobhans last updated on 16/Nov/21

$$\underset{x\to 0}{\lim }\frac{8(\sqrt{1+\tan {}^2\mathrm{x}}-1)+\tan {}^4\mathrm{x}-4\tan {}^2\mathrm{x}}{\tan {}^6\mathrm{x}}$$$$={\lim }_{\mathrm{x}\to 0}\frac{8(\sqrt{1+\mathrm{x}}-1)+{\mathrm{x}}^2-4\mathrm{x}}{{\mathrm{x}}^3};\mathrm{x}=\tan {}^2\mathrm{x}$$$$={\lim }_{\mathrm{x}\to 0}\frac{8(1+\frac{1}{2}\mathrm{x}-\frac{1}{8}{\mathrm{x}}^2+\frac{1}{16}{\mathrm{x}}^3-1)+{\mathrm{x}}^2-4\mathrm{x}}{{\mathrm{x}}^3}$$$$={\lim }_{\mathrm{x}\to 0}\frac{1}{2}=\frac{1}{2}$$

Answered by chhaythean last updated on 16/Nov/21

$$=\underset{x\to 0}{\lim }\frac{8\sec x-8+{\tan }^{4}x-{4\tan }^{2}x}{{\tan }^{6}x}\times \frac{{\tan }^{6}x}{x^6}$$$$=\underset{x\to 0}{\lim }\frac{8\sqrt{1+{\tan }^{2}x}-8+{\tan }^{4}x-{4\tan }^{2}x}{{\tan }^{6}x}$$$$\mathrm{let}\mathrm{u}={\tan }^{2}x,x\to 0\Rightarrow \mathrm{u}\to 0$$$$=\underset{\mathrm{u}\to 0}{\lim }\frac{8\sqrt{1+\mathrm{u}}-8+{\mathrm{u}}^2-4\mathrm{u}}{{\mathrm{u}}^3}$$$$=\underset{\mathrm{u}\to 0}{\lim }\frac{[8\sqrt{1+\mathrm{u}}-(8-{\mathrm{u}}^2+4\mathrm{u})][8\sqrt{1+\mathrm{u}}+(8-{\mathrm{u}}^2+4\mathrm{u})]}{{\mathrm{u}}^3[8\sqrt{1+\mathrm{u}}+(8-{\mathrm{u}}^2+4\mathrm{u})]}$$$$=\underset{\mathrm{u}\to 0}{\lim }\frac{64+64\mathrm{u}-{(8-{\mathrm{u}}^2+4\mathrm{u})}^2}{{\mathrm{u}}^3[8\sqrt{1+\mathrm{u}}+(8-{\mathrm{u}}^2+4\mathrm{u})]}$$$$=\underset{\mathrm{u}\to 0}{\lim }\frac{64+64\mathrm{u}-(64+{\mathrm{u}}^4+{16\mathrm{u}}^2-{16\mathrm{u}}^2-{8\mathrm{u}}^3+64\mathrm{u})}{{\mathrm{u}}^3[8\sqrt{1+\mathrm{u}}+(8-{\mathrm{u}}^2+4\mathrm{u})]}$$$$=\underset{\mathrm{u}\to 0}{\lim }\frac{-{\mathrm{u}}^4+{8\mathrm{u}}^3}{{\mathrm{u}}^3[8\sqrt{1+\mathrm{u}}+(8-{\mathrm{u}}^2+4\mathrm{u})]}$$$$=\underset{\mathrm{u}\to 0}{\lim }\frac{-\mathrm{u}+8}{[8\sqrt{1+\mathrm{u}}+(8-{\mathrm{u}}^2+4\mathrm{u})]}$$$$=\frac{8}{16}=\frac{1}{2}$$$$\mathrm{So}\overline{)\begin{array}{c}\underset{x\to 0}{\lim }\frac{8\sec x-8+{\tan }^{4}x-{4\tan }^{2}x}{x^6}=\frac{1}{2}\end{array}}$$

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