Question

Solve:


$\left(1\right)\underset{\mathrm{x}\to \mathrm{x}}{\mathrm{Lim}}\frac{1-{\cos }^3\mathrm{x}+{\sin }^3\mathrm{x}+\mathcal{l}\mathrm{n}\left(1+\cos \mathrm{x}\right)}{{\mathrm{x}}^2-1+2{\cos }^2\mathrm{x}+{\tan }^4\mathrm{x}+{\sin }^3\mathrm{x}}\mathrm{is}\mathrm{equal}\mathrm{to}-$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{1-{\cos }^{3}x+si{n}^{3}x+\ln \left(1+x^3\right)+\ln \left(1+\cos x\right)}{x^2-1+2{\cos }^{2}x+{\tan }^{4}x+{\sin }^{3}x} \\ Simplyputx=0intheexpression \\ =\frac{1-{\cos }^{3}0+si{n}^{3}0+\ln \left(1+0\right)+\ln \left(1+\cos 0\right)}{0-1+2{\cos }^{2}0+{\tan }^{4}0+{\sin }^{3}0} \\ =\frac{1-1+0+\ln 1+\ln \left(1+1\right)}{0-1+2+0+0} \\ =\frac{0+\ln 2}{1} \\ =\ln 2 \end{gathered}$$