limit does not exist??

S

shizzy

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Aug 10, 2005
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lim (cos(2) - cos(x))
x->infinity

Sorry don't know how to put the infinity sign in there. My question is that the text I am reviewing says that this limit does not exist and I don't understand why. I know the following about limits not existing:

A limit does not exist if:

1. The function is oscillating at/around the point the limit is being taken at
2. The function approaches a different number from the left(neg) than from the right(pos) side.
3. in general, a limit does not exist if evaluating the limit, it does not settle down on a single number.

Thanks for any help on this. I'm sure it's really easy and I'm just missing some basic thing!!
 
I thought about this some more and I think you are right, because, we can break up this problem into:

lim cos2 - lim cosx
x->infinity x->infinity

= cos2 - cos(infinity)

where we run into the problem that as you said, since cos varies from -1 to 1 (i think), we don't have any idea what value this will be at infinity. So in contrast if we just had:

lim x
x->infinity

for something like this the limit DOES exist since we know that as x approaches infinity, it(the limit) will equal infinity (whatever that is).

So...am I thinking correcly here or am I totally wrong?
 
Well, I don't want to sound like I know what I'm doing because I just decided to learn basic calculus over the summer, before school you know :)

So, the formula is basically:
eqn2186.png


Graphically, it looks like
T0_-1_562313.png


It would seem that the reason a limit doesn't exist — even at infinity, is because the function is constantly oscillating.
 
Hmmm... I'm not sure that's right because you CAN take the limit of cos(x). All we're doing when taking the limit is determining how a function is behaving at a specific point (although the text takes paintstaking effort to say that the limit is only interested in what is happening AROUND the point, not at it). So if I understand it correctly, if we take the limit of cos(x) at 0 for example, the limit L = 1, because the function at that point is approaching 1 from both the pos and neg side.

The example I have where the limit doesn't exist is f(x) = 1/cos x where the limit is taken at zero. Obviously this is undefined and hence the reason we're taking the limit at this point. The graph they have of this shows that the function looks normal except at x=0 where the graph oscillates VERY WILDLY around zero never settling on any given number, then goes back to normal past zero.

Just to clarify, the only reason I wrote all this is to show the tutors here what my understanding of this is so they can tell me if I'm right about this or if I have some flaw in my understanding. Everything I have written here could be ENTIRELY WRONG. Also note that I just wrote this very generally and vaguely. I do understand this in more depth( I think I do), I just didn't feel like articulating specifics.

So back to my original question, can anyone tell me the reason that the limit of cos(x) is not defined at infinity?
 
Actually, it's right. Think about it this way. A limit exists at only one y at a given point right? Let's say for your problem that as x approaches 3, the limit would be y= something.. Well, that's at one pt. But, if for example, x approaches 3 at -1 and 1, then there would be no limit right???

Kind of the same concept. At infinity, it oscilates. Only at infinity can this problem have no limit. Otherwise, it has a limit at every other x coordinate. :)...
 
Your problem is that you want to think of infinity as a "number" or a "point" but it isn't. It is just indefinitly large. Lim(x) as x-> infinity is NOT defined. Your #3 covers it. cos(x) just swings between -1 and +1 as x trundles along getting bigger and bigger.
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Gene
 
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