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Question Number 159646 by cortano last updated on 19/Nov/21

$$\underset{x\to 0}{\lim }\frac{1-{\left(\cos x\right)}^{\sin x}}{x^3}=?$$

Answered by FongXD last updated on 19/Nov/21

$$\mathrm{L}=\underset{\mathrm{x}\to 0}{\lim }\frac{1-{\left(\mathrm{cosx}\right)}^{\mathrm{sinx}}}{{\mathrm{x}}^3}\mathrm{form}:\frac{0}{0}$$$$\mathrm{L}=-\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{sinxln}\left(\mathrm{cosx}\right)}-1}{\mathrm{sinxln}\left(\mathrm{cosx}\right)}\times \frac{\mathrm{sinx}}{\mathrm{x}}\times \frac{\ln (1+\mathrm{cosx}-1)}{\mathrm{cosx}-1}\times \frac{1-\mathrm{cosx}}{{\mathrm{x}}^2}(-1)$$$$\mathrm{L}=-1\times 1\times 1\times \frac{1}{2}(-1)=\frac{1}{2}$$

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