Limits approaching infinity: sin(x)/x, cos(x), piecewise,...

[Rostron]

I am very unfamiliar with limits problems, and need to learn how to do them for this years upcoming calculus classes. From what I've read online, I understand the basics of it, but what about limits problems that are infinite? Here are a few example problems I as unable to figure out:

\(\displaystyle \lim_{x\to \infty} \frac{sinx}{x}\)

\(\displaystyle \lim_{x\to \infty} cosx\) This one just repeats a cycle, so what is it approaching?

\(\displaystyle \ h(x) = \left\{ \begin{array}{ll}
sinx & \mbox{if x < \frac{\pi}{2}}\\
1 & \mbox{if \frac{\pi}{2} \leq x \leq 2 }\\
1-x & \mbox{if x > 2}\end{array} \right. \\) this corresponds with the below problems and has me absolutely lost 100%

a. \(\displaystyle \lim_{x\to \frac{\pi}{2}} h(x)\)

b. \(\displaystyle \lim_{x\to 2} h(x)\)

c. \(\displaystyle \lim_{x\to \frac{-\pi}{2}} h(x)\)

I'm very new with limits so forgive me if these problems actually aren't difficult.

 

[pka]

Do you understand that \(\displaystyle \lim _{x \to \infty } \frac{1}{x} = 0\)?
If so, the function \(\displaystyle \sin (x)\) bounded, therefore \(\displaystyle \lim _{x \to \infty } \frac{{\sin (x)}}{x} = 0\).

For the \(\displaystyle h(x)\) problem, consider what happens on the left and right hands.
For example: \(\displaystyle \lim _{x \to 2^ - } h(x) = 1\) so \(\displaystyle \lim _{x \to 2^ + } h(x) = ?\)

 

[Rostron]

Thanks pka, I understand the sin(x)/x limit. For the cos(x) problem, would the answer just be f(x)=cosx?
The h(x) problem is much more confusing...if h(x)=1, wouldn't it just be approaching 2 from the left and pi/2 from the right?

 

[tkhunny]

1) Fix a technical concept and your life will be easier and your understanding increased.

"approaches infinity" doesn't mean anything. Don't say it, write it, or think it. The term you want to use is "increases without bound". Infinity isn't anywhere to approach, so don't try to do it.

2) It may seem a little obvious, but in order to approach something, something must be approached. You are right to be suspicopus about cos(x) as x increases without bound. cos(x) just keeps jumping around. There is no limit.