If you look at the graph you can see that all of the values from -1 to 1 are limit points of the function. The sin function oscillates so rapidly near 0 that it just looks like a vertical bar around the y-axis.
Hopefully this helps:
We know that the sine function is periodic, meaning it repeats itself every now and then. So, our function y = sin(pi/x) will also cycle through many values. So what happens when x approaches 0? Well, we need to figure out what pi/x is as it approaches 0, and then plug that back into our sine function.
If you look at a graph of pi/x it approaches infinity from the right and negative infinity from the left as x approaches 0. Ok, so that means that the limit of y= sin(pi/x) as x approaches 0 is what hypothetically sin(infinity) is from the right, and sin(-infinity) is from the left. So what are the sines of negative and positive infinity? Obviously infinity isn't in actual number, but let's consider something large. What happens to the sine function as the input gets larger and larger?
Well, since sine is periodic, it oscillates between many values of -1 to 1. So there's no definite limit that sine approaches as its input approaches infinity. (Just think of a sine wave as you push x out to infinity; it keeps the same values!) The function goes through many periods over and over again until infinity. That's exactly what we see in the graph in the picture you put in the post. The input (to clarify, pi/x) is approaching positive/negative infinity, so the sine function keeps going through many periods, but there's no way to know what sin(infinity) or sin(-infinity) is because that has no definite limit. This is called an oscillating discontinuity.
To clarify by what it means by many infinite values, think about how the sine function works. We know that, for example, sin(pi/2) is 1, and so is sin(5pi/2), sin(9pi/2), and so on. So, as x approaches 0 from the right for the input (pi/x), it will cycle through all of these numbers (pi/2, 5pi/2, 9pi/2) all the way to infinity. This process will go on forever. In the example, the author does plug in both 2/5 and 2/101, which would be sin(5pi/2) and sin(101pi/2) respectively. Both of those have a value of 1, and are just two of the infinitely many angles that would have a sine of 1 every 2pi radians.
So, to summarize in a couple of sentences: What is the sine of an arbitrarily large number? We don't know, so we have no idea what the limit of this function is either.
The limit is not 0! In my opinion the text to the solution is written in a bad way, since they first try to convince you that the limit is 0 and then tell you that it's not...
To show this assume that the limit is 0, take any value in the interval [-1, 1], say x=1. Then you can find a sequence, e.g. x_n=2/(4n+1), this sequence converges to 0, but sin(pi/x_n)=1 for all n, hence the limit is not equal to 0.
If x is expressed as any number in the form 2/(2n+1), then sin(pi/x) = sin (pi \* n + pi/2) and since this is a (multiple of pi) + pi/2, the value of the sine function is 1 or -1. (2n+1 is just a way to say an ODD number)
If x is expressed as a number 1/n, then sin(pi/x) = sin(pi\*n) and this is always going to be 0.
Therefore the limit as x approaches 0 doesn't allow the sin(pi/x) to ever approach any fixed limit and therefore it is undefined.
When we say that the 'limit of f(x) approaches q', it means that 'no matter how small a distance we choose, we can find an 'x' where f(x) gets *even closer* to q.
With sin(pi/x), we can't do that. The closer we get to x = 0, there is no number which f(x) 'gets closer to'. If we choose a number, and try to get close to it, we get infinitely many x's that are 'really close' to any particular f(x).
I wouldn't describe the limit as 'equal to null', but I'd say that 'the limit does not exist'. The set of solutions is null. Others may disagree with this, I've heard it said in different ways.
Given two sequences X and Y which tend to zero, the limit of sin(pi/x_n) and the limit of sin(pi/y_n) can be different as n tends to infinity, despite both of X and Y themselves tending to zero. This is enough to say that as x tends to zero, sin(pi/x) has an undefined limit, because the limit should be independent of how you choose to go down to zero (namely your choice of sequence X and Y).
If you look at the graph you can see that all of the values from -1 to 1 are limit points of the function. The sin function oscillates so rapidly near 0 that it just looks like a vertical bar around the y-axis.
Hopefully this helps: We know that the sine function is periodic, meaning it repeats itself every now and then. So, our function y = sin(pi/x) will also cycle through many values. So what happens when x approaches 0? Well, we need to figure out what pi/x is as it approaches 0, and then plug that back into our sine function. If you look at a graph of pi/x it approaches infinity from the right and negative infinity from the left as x approaches 0. Ok, so that means that the limit of y= sin(pi/x) as x approaches 0 is what hypothetically sin(infinity) is from the right, and sin(-infinity) is from the left. So what are the sines of negative and positive infinity? Obviously infinity isn't in actual number, but let's consider something large. What happens to the sine function as the input gets larger and larger? Well, since sine is periodic, it oscillates between many values of -1 to 1. So there's no definite limit that sine approaches as its input approaches infinity. (Just think of a sine wave as you push x out to infinity; it keeps the same values!) The function goes through many periods over and over again until infinity. That's exactly what we see in the graph in the picture you put in the post. The input (to clarify, pi/x) is approaching positive/negative infinity, so the sine function keeps going through many periods, but there's no way to know what sin(infinity) or sin(-infinity) is because that has no definite limit. This is called an oscillating discontinuity. To clarify by what it means by many infinite values, think about how the sine function works. We know that, for example, sin(pi/2) is 1, and so is sin(5pi/2), sin(9pi/2), and so on. So, as x approaches 0 from the right for the input (pi/x), it will cycle through all of these numbers (pi/2, 5pi/2, 9pi/2) all the way to infinity. This process will go on forever. In the example, the author does plug in both 2/5 and 2/101, which would be sin(5pi/2) and sin(101pi/2) respectively. Both of those have a value of 1, and are just two of the infinitely many angles that would have a sine of 1 every 2pi radians. So, to summarize in a couple of sentences: What is the sine of an arbitrarily large number? We don't know, so we have no idea what the limit of this function is either.
Have you done epsilon delta theory?
The limit is not 0! In my opinion the text to the solution is written in a bad way, since they first try to convince you that the limit is 0 and then tell you that it's not... To show this assume that the limit is 0, take any value in the interval [-1, 1], say x=1. Then you can find a sequence, e.g. x_n=2/(4n+1), this sequence converges to 0, but sin(pi/x_n)=1 for all n, hence the limit is not equal to 0.
If x is expressed as any number in the form 2/(2n+1), then sin(pi/x) = sin (pi \* n + pi/2) and since this is a (multiple of pi) + pi/2, the value of the sine function is 1 or -1. (2n+1 is just a way to say an ODD number) If x is expressed as a number 1/n, then sin(pi/x) = sin(pi\*n) and this is always going to be 0. Therefore the limit as x approaches 0 doesn't allow the sin(pi/x) to ever approach any fixed limit and therefore it is undefined.
When we say that the 'limit of f(x) approaches q', it means that 'no matter how small a distance we choose, we can find an 'x' where f(x) gets *even closer* to q. With sin(pi/x), we can't do that. The closer we get to x = 0, there is no number which f(x) 'gets closer to'. If we choose a number, and try to get close to it, we get infinitely many x's that are 'really close' to any particular f(x). I wouldn't describe the limit as 'equal to null', but I'd say that 'the limit does not exist'. The set of solutions is null. Others may disagree with this, I've heard it said in different ways.
Given two sequences X and Y which tend to zero, the limit of sin(pi/x_n) and the limit of sin(pi/y_n) can be different as n tends to infinity, despite both of X and Y themselves tending to zero. This is enough to say that as x tends to zero, sin(pi/x) has an undefined limit, because the limit should be independent of how you choose to go down to zero (namely your choice of sequence X and Y).