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Question Number 137157 by mathlove last updated on 30/Mar/21

Commented by mathlove last updated on 30/Mar/21

$$withoutmacloreensireesandH-pitalrools$$

Commented by mathlove last updated on 30/Mar/21

Commented by bobhans last updated on 30/Mar/21

$${\mathrm{L}}^{\prime }\mathrm{Hopital}$$$$\underset{x\to 0}{\lim }\frac{1-\cos \mathrm{x}}{1-\sec {}^2\mathrm{x}}=\underset{x\to 0}{\lim }\frac{\cos {}^2\mathrm{x}(1-\cos \mathrm{x})}{(\cos \mathrm{x}-1)(\cos \mathrm{x}+1)}$$$$=-\frac{1}{2}$$

Commented by mathlove last updated on 30/Mar/21

$$wethoutLHopitalrols$$

Commented by bobhans last updated on 30/Mar/21

$$\underset{x\to 0}{\lim }\frac{\mathrm{x}-\sin \mathrm{x}}{{\mathrm{x}}^3}.\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^3}{\mathrm{x}-\tan \mathrm{x}}$$$$\mathrm{you}\mathrm{can}\mathrm{solve}\mathrm{it}\mathrm{without}{\mathrm{L}}^{\prime }\mathrm{Hopital}$$

Commented by mathlove last updated on 30/Mar/21

$$noIcannot$$

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