Sunday, October 9, 2016

limits - Avoid L'hopital's rule





$$\lim_{x\to 0} {\ln(\cos x)\over \sin^2x} = ?$$



I can solve this by using L'Hopital's rule but how would I do this without this?


Answer



$$\frac{\log\left(\cos\left(x\right)\right)}{\sin^{2}\left(x\right)}=\frac{1}{2}\frac{\log\left(1-\sin^{2}\left(x\right)\right)}{\sin^{2}\left(x\right)}=-\frac{\sin^{2}\left(x\right)+O\left(\sin^{4}\left(x\right)\right)
}{2\sin^{2}\left(x\right)}\stackrel{x\rightarrow0}{\rightarrow}-\frac{1}{2}.$$



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