Can somebody explain why the function sin 1/x has no limit, or any oscillating function in general?

Question

please keep in mind I am a beginning calculus student. Telling me why limits are important would help too, but I sort of get that stuff, I just don't know why the function sin 1/x limit isn't 0.

Answer 1

The only place where f(x) = sin (1/x) doesn't have a limit is x=0.

The reason it doesn't is: Look at the interval [-d,d] for ANY positive d. sin (1/x) takes all values in the range [-1,1] for various values of x in that interval. So if you set epsilon = 1/2, say, it is NOT the case that there exists a delta such that ... well, just look up the definition of "limit" to complete the sentence.

A function is continuous/has a limit at a point when, as x gets near that point, the value of the function settles down to increasingly small ranges about the limit. At x=0, sin(1/x) does NOT have that property.

Answer 2

I'll first deal with the reason why the function sin 1/x has no limit AS x ---> 0. [You didn't SPECIFY the value of x ! I PRESUME you meant at x = 0, since sin (1/x) is of course perfectly continuous for all x not equal to 0.]

That assumed, here is why there is no limit to sin(1/x) as x ---> 0 :

Because, as you apparently know, the definition of a limit tells you that given a (small, positive) epsilon there is always a delta such that for x within delta of the limit value, x_0, the VALUE of the function is always within +/- epsilon of that quantity [ f(x_0) ] defined as its limit as x ---> x_0.

That simply can't apply to sin (1/x) as x ---> 0 ! Why not? Because no matter HOW CLOSE x gets to 0, there will always be an INFINITE NUMBER of x-values CLOSER to 0 where the function takes ANY VALUE WHATSOEVER between -1 and +1. Putting it into more colloquial language, you simply cannot identify one particular value that sin (1/x) will be trapped near, as x tends to zero.

The same argument DOESN'T APPLY to "any oscillating function in general." You have to examine them on a case by case basis. (You're not being specific enough: sin x, cos x are "oscillating functions"). I think that what you meant is "to oscillating functions that oscillate infinitely with decreasing distances between nodes as some particular value of x is approached."

But in fact even that isn't true. For example, the functions:

F(x) = x sin (1/x) or G(x) = x^2 sin (1/x) etc. DO have a LIMIT, 0 at x = 0. That is because, while the sine part still oscillates wildly as as
x ---> 0, its value only lies between - 1 and + 1. Meanwhile, the factors x , x^2, ... etc. are still effective in "trapping" F(x), G(x). etc. closer and closer to zero.

On the other hand,

H(x) = [ sin(1/x) ] / x^n for ALL positive n inherits the combined pathologies of both sin(1/x) AND 1/x^n as x ---> 0; no limit therefore exists.

Note that I've only used sin(1/x) in these examples since you're familiar with it. But any function that has some wave-like ("oscillating") character to its values for arbitrarily large x will behave pathologically near x = 0 if you switch the argument from x to (1/x). They will thus have no limit as x ---> 0, just as the original function had no limit as x ---> "infinity." That's just in the nature of the beast!

I hope that this has explained why it's not only difficult but also impossible to state a hard and fast rule about "oscillating functions in general."

Live long and prosper.

Answer 3

In this case you expand the function and apply the limit It looks like the limit is 0 by my memory. You have to expand (x-1)^-2 apply the limit

Answer 4

First of all, whenever talking about limit of f(x) you need to specify what x is approaching.

I assume that here, x is approaching 0. (because the limit of sin (1/x) as x approaches infinity is 0)..

Consider the graph of sin(x). It's like a wave, right? it goes up to 1 then down through 0 and to -1 and then back up continuously. as x approaches infinity, y goes to 1 then -1 (no limit)

Let's plug number in just for fun!
sin(99) = - .9992
sin(9999) = .6361
sin(999999) = - .97735
This pattern continues: positive, negative, etc.
Therefore limit of sin(x) as x approaches infinity is undefined--no limit!

How does this help?
1/x as x approaches infinity is 0. (As x becomes really big, 1/x becomes really small)
1/x as x approaches 0 is infinity. (As x becomes really small, 1/x becomes really big!)

So
Limit sin(x) as x approaches infinity equals sin(1/x) as x approaches 0.

Hope this helps.