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Question Number 167795 by niverland last updated on 25/Mar/22

Commented by MJS_new last updated on 25/Mar/22

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{the}\mathrm{limit}\mathrm{exists}.\mathrm{I}\mathrm{get}$$$$\left(1\right)\underset{x\to 0^{-}}{\lim }(...)=+\mathrm{\infty }$$$$\left(2\right)\underset{x\to 0^{+}}{\lim }(...)=-\mathrm{\infty }$$

Answered by qaz last updated on 25/Mar/22

$$\underset{\mathrm{x}\to 0}{\lim }\frac{\tan ({\mathrm{e}}^{\sin \mathrm{x}}-\sin -1)-\ln (2-\cos \mathrm{x})}{\sqrt[5]{1-{\mathrm{x}}^3}-1}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{\sec {}^2({\mathrm{e}}^{\sin \mathrm{x}}-\sin \mathrm{x}-1)\cdot ({\mathrm{e}}^{\sin \mathrm{x}}\cos \mathrm{x}-\cos \mathrm{x})-\frac{\sin \mathrm{x}}{2-\cos \mathrm{x}}}{{(-\frac{1}{5}{\mathrm{x}}^3)}^{\prime }}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{(2-\cos \mathrm{x})\cdot \cos \mathrm{x}\cdot \sec {}^2({\mathrm{e}}^{\sin \mathrm{x}}-\sin \mathrm{x}-1)\cdot ({\mathrm{e}}^{\sin \mathrm{x}}-1)-\sin \mathrm{x}}{-\frac{3}{5}{\mathrm{x}}^2(2-\cos \mathrm{x})}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{(2-\cos \mathrm{x})\cdot \cos \mathrm{x}\cdot \sec {}^2({\mathrm{e}}^{\sin \mathrm{x}}-\sin \mathrm{x}-1)\cdot (\sin \mathrm{x}+\frac{1}{2}\sin {}^2\mathrm{x}+\mathrm{o}\left({\sin }^2\mathrm{x}\right))-\sin \mathrm{x}}{-\frac{3}{5}{\mathrm{x}}^2(2-\cos \mathrm{x})}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{(2-\cos \mathrm{x})\cdot \cos \mathrm{x}\cdot \sec {}^2({\mathrm{e}}^{\sin \mathrm{x}}-\sin \mathrm{x}-1)\cdot (1+\frac{1}{2}\sin \mathrm{x}+\mathrm{o}\left(\sin \mathrm{x}\right))-1}{-\frac{3}{5}\mathrm{x}(2-\cos \mathrm{x})}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{\ln [(2-\cos \mathrm{x})\cdot \cos \mathrm{x}\cdot \sec {}^2({\mathrm{e}}^{\sin \mathrm{x}}-\sin \mathrm{x}-1)\cdot (1+\frac{1}{2}\sin \mathrm{x}+\mathrm{o}\left(\sin \mathrm{x}\right))]}{-\frac{3}{5}\mathrm{x}(2-\cos \mathrm{x})}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{\ln (1+\frac{1}{2}\sin \mathrm{x}+\mathrm{o}\left(\sin \mathrm{x}\right))}{-\frac{3}{5}\mathrm{x}(2-\cos \mathrm{x})}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{\frac{1}{2}\sin \mathrm{x}+\mathrm{o}\left(\sin \mathrm{x}\right)}{-\frac{3}{5}\mathrm{x}(2-\cos \mathrm{x})}$$$$=-\frac{5}{6}$$$$-----------$$$$\underset{\mathrm{x}\to 0}{\lim }\frac{\ln \left(\sec {}^2({\mathrm{e}}^{\sin \mathrm{x}}-\sin \mathrm{x}-1)\right)}{-\frac{3}{5}\mathrm{x}(2-\cos \mathrm{x})}=\underset{\mathrm{x}\to 0}{\lim }\frac{-2\ln \left(\cos ({\mathrm{e}}^{\sin \mathrm{x}}-\sin \mathrm{x}-1)\right)}{-\frac{3}{5}\mathrm{x}(2-\cos \mathrm{x})}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{-2(\cos (\frac{1}{2}\sin {}^2\mathrm{x}+\mathrm{o}\left(\sin {}^2\mathrm{x}\right))-1)}{-\frac{3}{5}\mathrm{x}(2-\cos \mathrm{x})}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{-2(-\frac{1}{8}\sin {}^4\mathrm{x}+\mathrm{o}\left(\sin {}^4\mathrm{x}\right))}{-\frac{3}{5}\mathrm{x}(2-\cos \mathrm{x})}=0$$

Commented by MJS_new last updated on 25/Mar/22

$$f\left(x\right)=\frac{\tan ({\mathrm{e}}^{\sin x}-\sin x-1)+\ln (2-\cos x)}{\sqrt[5]{1-x^3}-1}$$$$f(0.1)\approx -50.6640$$$$f(0.01)\approx -500.819$$$$f(0.001)\approx -5000.83$$$$f(0.0001)\approx -50001.0$$$$...$$$$f(-0.1)\approx 49.0447$$$$f(-0.01)\approx 499.152$$$$f(-0.001)\approx 4999.17$$$$f(-0.0001)\approx 49999.0$$$$...$$

Answered by qaz last updated on 25/Mar/22

$$\underset{\mathrm{x}\to 0}{\lim }\frac{\tan ({\mathrm{e}}^{\sin \mathrm{x}}-\sin \mathrm{x}-1)-\ln (2-\cos \mathrm{x})}{\sqrt[5]{1-{\mathrm{x}}^3}-1}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{({\mathrm{e}}^{\sin \mathrm{x}}-\sin \mathrm{x}-1+...)-(1-\cos \mathrm{x}+...)}{-\frac{1}{5}{\mathrm{x}}^3}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{(\frac{1}{2}\sin {}^2\mathrm{x}+\frac{1}{6}\sin {}^3\mathrm{x}+...)-(\frac{1}{2}{\mathrm{x}}^2+...)}{-\frac{1}{5}{\mathrm{x}}^3}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{\frac{1}{2}{\mathrm{x}}^2{(1+...)}^2+\frac{1}{6}{\mathrm{x}}^3{(1+...)}^3-\frac{1}{2}{\mathrm{x}}^2+...}{-\frac{1}{5}{\mathrm{x}}^3}$$$$=\underset{\mathrm{x}\to 0}{\lim }\frac{\frac{1}{6}{\mathrm{x}}^3+\mathrm{o}\left({\mathrm{x}}^3\right)}{-\frac{1}{5}{\mathrm{x}}^3}$$$$=-\frac{5}{6}$$

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