Examine the continuity at 0 of f such that f(x) = sin(1/x)?

Question

Examine the continuity at 0 of f such that f(x) = sin(1/x) when x not= 0 and f(0) = 0

Thnx
GTHIRU

Answer 1

The sin function always produces a value between -1 and 1, so the function will not go off to infinity.

sin(1/x) is a very cool function in my opinion. As it approaches 0 from either side, it oscillates between -1 and 1 faster and faster until it becomes a vertical blur very close to 0. Not only does the two-sided limit not exist, but the one-sided limits don't exist either.

Let's consider the limit from the right (the left side is analogous). Pick an extremely large value for x, so that 1/x is extremely small. Now there are always two numbers greater than x, call them x1 and x2, such that sin(x1) = 1 and sin(x2) = -1. Since 1/x1 and 1/x2 are smaller than x, and our choice of x was arbitrary, the function sin(1/x) cannot have a limit as x approaches 0.

The function is interesting precisely because it oscillates infinitely many times on each side of zero. It's an example of a curve with infinite length that fits into a very small piece of the xy-plane!

Answer 2

f(x) = sin(x), as you know goes back and forth between 1 and -1 in regular intervals.

As f(x) = sin(1/x) goes to zero, it goes back and forth between 1 and -1 with ever increasing frequency. So within any small interval, there are still infinitely many values of x such that f(x) = 1, and there are still infinitely many values of x such that f(x) = -1,and there are still infinitely many values of x such that f(x) =any value you want to choose between 1 and -1.

This means that f(x) = sin(1/x) does not converge to any particular value as x goes to 0. Since the function is divergent at x = 0, it cannot be continuous.

I hope this is clear!

Answer 3

Actually, the limit of sin(1/x) _does_not_exist_ as x approaches zero from either the left or right (it's not infinity as the previous answerer posted... sine never gets to infinity... it's sorta hopelessly constrained between -1 and 1)
Since one definition of continuity is that lim as x approaches a for f(x) = f(a), it follows that we do not have continuity, since the limit does not exist.

It follows, methinks, that sin(1/x) is discontinuous at x=0

Answer 4

As x --> 0 from the right, you have a sine curve that is getting really squished in on itself, but never takes on any vaue outside the range [-1,1].

In fact, sin(1/x) is continuous on R - {0} (all real numbers except 0).

Although sin(1/x) is not continuous in and of itself, the set "sin(1/x) unioned with {0}x[-1,1]" (the closed set [-1,1] on the y-axis) is continuous. I would have to dig out my topology book to give you the proof...but it is at the house.

Answer 5

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