Explain why sin(1/x) oscillates infinitely as x approaches 0.?

Answer 1

Sin x is periodic with period 2pi. It takes ever smaller changes in x to get 1/x to change by 2pi.

Sketch y=1/x then mark off equal steps along y. Note that the closer you get to x=0 the smaller the x steps are to yield the fixed y steps, so sin 1/x repeats at ever smaller steps in x as x goes to 0.

Answer 2

Remember that sin(q) goes up and down every time q changes by pi=3.14159...So the more that q changes, the more times sin(q) goes up and down.

As x approaches 0, 1/x goes up, really fast. Really, really fast. So sin(1/x) goes up and down (oscillates) really, really, fast, too.

Example. If x = 0.001, 1/x = 1000. If x = 0.002, 1/x = 500. So between 0.001 and 0.002, 1/x changes by 500, meaning over 100 oscillations.

By the way, sin(1/x) oscillates slower as x gets higher. In fact, the last full oscillation starts at about x=1/6 (it's actually 1/(2*pi)), and "ends" at positive infinity. And that's just the one oscillation.

Answer 3

sin (1/x) has no limit therefore it does not approach a single number as you approach 0. This means it will continue to oscillate more and more as you approach 0.

Answer 4

Because x can never equal zero (1/0 is undefined).