Consider the function f(x)=sin(1/x)
Find a sequence of x-values that approach 0 such that
sin (1/x)=0
sin (1/x)=1
sin (1/x)=-1
Is sin sin (1/x)=0 and sin (1/x)=-1 does not exist.
What is sin (1/x)=1 then.
sin (1/x) = 0 if 1/x = pi, which means x = 1/pi
sin (1/x) = 1 if 1/x = pi/2, which means x = 2/pi
sin (1/x ) = -1 if 1/x = 3 pi/2, which means x = 2/(3 pi)
I don't understand why the x values should approach zero, or what the rules of the sequence are.
To find a sequence of x-values that approach 0 such that sin(1/x) = 1, we can first note that the sine function takes values between -1 and 1. The value sin(1/x) = 1 corresponds to the point at which the graph of sin(1/x) intersects the line y = 1.
To find this intersection point, we can set sin(1/x) equal to 1:
sin(1/x) = 1
Next, we can take the inverse sine of both sides to find the angle that has a sine of 1:
arcsin(sin(1/x)) = arcsin(1)
Simplifying, we have:
1/x = π/2
Solving for x, we get:
x = 2/π, 4/π, 6/π, 8/π, ...
So, the sequence of x-values that approach 0 such that sin(1/x) = 1 is {2/π, 4/π, 6/π, 8/π, ...}.