Question

How do I prove that this function is NOT bounded?

Answer 1

What is the funtion?

Answer 2

Here

Answer 3

By the graph you can tell it is not bounded but how to prove it? :S

Answer 4

it is unbounded because \(2x\sin(\frac{1}{x^2})\) is bounded by \(2x\) but \[\frac{\cos(\frac{1}{x^2})}{x^2}\] has a numerator between -1 and 1, but a denominator that goes to zero

Answer 5

@satellite73 Sorry, I don't get it. Could this be an "acceptable" answer in a formal test?

I mean how do I prove that 2x sin(1/x^2) is bounded?

And also cos(1/x^2)/x^2 fluctuates between -oo and +oo

Answer 6

what class is this for?

Answer 7

advanced calc/analysis?

Answer 8

It's calculus 1

Answer 9

do you get that 2xsin(1/x^2) is bounded by 2x as @satellite73 said?

Answer 10

note that -1<= sin(1/x^2) <= 1

Answer 11

hello?

Answer 12

Ye, I'm here , I can understand that it is bounded