limit to infinity and cos

G

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I'm having trouble with one of my questions. I need to find the limit(or find that it does not exist). The question is:
lim(t-> infinity) cos(t+5t^(-2))

Plugging in infinity gets me to cos(infinity), which doesn't really help me much, so I've been trying to get the equation to a format that I can use with little success. I don't think I can use the squeeze theorem, either(but please correct me if I'm wrong).

so far:
f(t)= cos(t+5(1/(t^2)))
=cos(t+(5/(t^2)))
But I can't get anything useful beyond that, as if I try to combine the t and 5/(t^2) and divide by the greatest power of t in the denominator, it leads me full circle.

Any help would be appreciated.
 
o.k. answer a more simple problem ...

does the following limit exist ?

\(\displaystyle \L \lim_{x \rightarrow \infty} \cos{(x)}\)

why or why not?
 
No, since cos graphs go up and down between -1 and 1 throughout the graph. It won't be closer to either value or to a value in between at infinity.

And replacing x with t+5t^-2 would get us the same answer to the previous question, correct? (although that would do strange things to the x->infinity , wouldn't it?)
 
bingo.

however, no strange behavior ... cos(t + 5/t<sup>2</sup>) would behave pretty much the same as cos(x) as both x and t -> infinity.
why?
 
I figured stuff out this morning. It doesn't matter what's inside the cos, because if there's nothing outside, then you know it will just keep going up and down forever between the same numbers.

I was just thinking that if you substituted t+5/t^2 for the x, then x->infinity would become t+5/t^2 ->infinity. Which, of course, would mess things up quite royally, except that that doesn't happen.

Thank you for the assistance!
 
It doesn't matter what's inside the cos ...
Click to expand...

careful ... what is the value of this limit?

\(\displaystyle \L \lim_{t\rightarrow \infty} \cos{(\frac{\pi t}{2t+1})}\)
 
Inside the cos can be simplified. It came out as cos pi/(2+(1/t))
Running that by the limit comes out as cos pi/2 = 0

It's because of dividing by the 2t+1, isn't it?
 
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