Find:
lim (lncosx)/x^2
x-->0
$\displaystyle ln(cos(0))=ln(1)$ so the logarithm with base e that equals 1 is just equal to zero. This means that the numerator and the denominator approach zero, so use l-hopitals rule.
$\displaystyle \lim_{x->0}\frac{ln(\cos(x))}{x^2}=\lim_{x->0}\frac{-\sin(x)}{2x\cos(x)}$
No just apply the rule a second time. It's pretty easy from here.
Notice that for $\displaystyle x$ within a sufficently small neighborhood of $\displaystyle 0$ that $\displaystyle \ln\left(\cos(x)\right)=\ln\left(\sqrt{1-\sin^2(x)}\right)=\frac{1}{2}\ln\left(1-\sin^2(x)\right)$. Also, it is a well known fact that $\displaystyle \sin(x)\underset{x\to0}{\sim}x$ and $\displaystyle \ln\left(1-x\right)\underset{x\to0}{\sim}-x$. So combining these two we can see that $\displaystyle \ln\left(1-\sin^2(x)\right)\underset{x\to0}{\sim}\ln\left(1-x^2\right)\underset{x\to0}{\sim}-x^2$. So we may conclude that $\displaystyle \lim_{x\to0}\frac{\ln\left(\cos(x)\right)}{x^2}=\l im_{x\to0}\frac{\tfrac{1}{2}\left(-x^2\right)}{x^2}=\frac{-1}{2}$.