1. ## Limits

find the limit as x-> inf for f(x)= cos(x)/sqrt(x)

so, I can tell it's going to be 0, but as a proof i'm not sure what to say

I started by changing it to sqrt(x)*cos(x)/x <= sqrt(x)/(x) = 1/sqrt(x)

Which obviously goes to zerp as x goes to infinity...but what about a lower bound?

Another problem is lim x -> 0 for f(x) = sin^2(sin(x))

All i've really got is that it is bounded below by 0 and above by 1, not really sure if a epsilon delta proof would be better...

2. Originally Posted by pkr
find the limit as x-> inf for f(x)= cos(x)/sqrt(x)

so, I can tell it's going to be 0, but as a proof i'm not sure what to say

I started by changing it to sqrt(x)*cos(x)/x <= sqrt(x)/(x) = 1/sqrt(x)

Which obviously goes to zerp as x goes to infinity...but what about a lower bound?
$\lim_{x\to\infty}\frac{-1}{\sqrt{x}}\leq\lim_{x\to\infty}\frac{\cos(x)}{\s qrt{x}} \leq \lim_{x\to\infty}\frac{1}{\sqrt{x}}$

Thus $0\leq\lim_{x\to\infty}\frac{\cos(x)}{\sqrt{x}}\leq 0$, so the limit equals zero by the squeeze theorem.

3. Originally Posted by pkr
Another problem is lim x -> 0 for f(x) = sin^2(sin(x))

All i've really got is that it is bounded below by 0 and above by 1, not really sure if a epsilon delta proof would be better...
$f(x)=\sin^2(\sin(x))$

This function is continuous, as it is the composition of two continuous functions, and thus $\lim_{x\to a}f(x) = f(a)$

Using $a=0$, we have $\lim_{x\to0}\sin^2(\sin(x)) = \sin^2(\sin(0)) = 0$