1. ## Limits(trigonometric)

Find:
lim (lncosx)/x^2
x-->0

2. Originally Posted by matsci0000
Find:
lim (lncosx)/x^2
x-->0

When the x-->0 the function is 0/0.
So use L'hopitals rule.
For this function, use it twice.

3. Originally Posted by matsci0000
Find:
lim (lncosx)/x^2
x-->0
$ln(cos(0))=ln(1)$ 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.

$\lim_{x->0}\frac{ln(\cos(x))}{x^2}=\lim_{x->0}\frac{-\sin(x)}{2x\cos(x)}$

No just apply the rule a second time. It's pretty easy from here.

4. Originally Posted by matsci0000
Find:
lim (lncosx)/x^2
x-->0
Notice that for $x$ within a sufficently small neighborhood of $0$ that $\ln\left(\cos(x)\right)=\ln\left(\sqrt{1-\sin^2(x)}\right)=\frac{1}{2}\ln\left(1-\sin^2(x)\right)$. Also, it is a well known fact that $\sin(x)\underset{x\to0}{\sim}x$ and $\ln\left(1-x\right)\underset{x\to0}{\sim}-x$. So combining these two we can see that $\ln\left(1-\sin^2(x)\right)\underset{x\to0}{\sim}\ln\left(1-x^2\right)\underset{x\to0}{\sim}-x^2$. So we may conclude that $\lim_{x\to0}\frac{\ln\left(\cos(x)\right)}{x^2}=\l im_{x\to0}\frac{\tfrac{1}{2}\left(-x^2\right)}{x^2}=\frac{-1}{2}$.