Consider the function f(x)=sin(1/x)
Find a sequence of x-values that approach 0 such that
(1) sin (1/x)=0 {Hint: Use the fact that sin(pi) = sin(2pi)=sin(3pi)=...=sin(npi)=0}
(2) sin (1/x)=1 {Hint: Use the fact that sin(npi)/2)=1 if n= 1,5,9...}
(3) sin (1/x)=-1
(4) Explain why your answers to any of parts(1-3) show that lim X->0 sin(1/x) does not exist.
Is sin sin (1/x)=0 and sin (1/x)=-1 does not exist.
What is sin (1/x)=1 then.
To find x-values that approach 0 and satisfy the given conditions, we can use the hints given. Let's go through each part one by one:
(1) To find x-values such that sin(1/x) = 0, we can set the argument of the sine function equal to multiples of pi. The argument 1/x should be equal to n * pi, where n is any integer except 0. So, we can write:
1/x = n * pi
Solving for x, we get:
x = 1/(n * pi)
Thus, the sequence of x-values that approach 0 and make sin(1/x) = 0 is x = 1/(n * pi) for any n ≠ 0.
(2) To find x-values such that sin(1/x) = 1, we use the fact that sin(n * pi/2) = 1 when n is odd. So, we can set the argument of the sine function equal to odd multiples of pi/2. The argument 1/x should be equal to (2n + 1) * pi/2, where n is any integer. Solving for x, we get:
x = 1/((2n + 1) * pi/2)
Thus, the sequence of x-values that approach 0 and make sin(1/x) = 1 is x = 1/((2n + 1) * pi/2) for any integer n.
(3) To find x-values such that sin(1/x) = -1, we use the fact that sin(n * pi/2) = -1 when n is even. So, we can set the argument of the sine function equal to even multiples of pi/2. The argument 1/x should be equal to 2n * pi/2, where n is any integer. Solving for x, we get:
x = 1/(2n * pi/2)
Thus, the sequence of x-values that approach 0 and make sin(1/x) = -1 is x = 1/(2n * pi/2) for any integer n.
Now, as for the explanation of why the limit of sin(1/x) as x approaches 0 does not exist, we can observe that the sine function oscillates infinitely between -1 and 1 as its argument approaches 0. This means that no matter how close we get to 0, sin(1/x) will take on infinitely many different values. Therefore, the limit of sin(1/x) as x approaches 0 does not exist.