Question

(Calculus 1) Help!
Prove that the following function is NOT bounded

Answer 1

Function attached below

Answer 2

oops correction: it's 2x sin(1/x^2)

Answer 3

first off, what does not bounded mean?

Answer 4

that f(x) can go as high and as low as possible

Answer 5

so no restrictions right?

Answer 6

a bounded function is |f(x)|<M ,M is the bound

Answer 7

Okay, so basically you have to prove that it has no restrictions then.

Answer 8

What do you mean by no restrictions?

Answer 9

bounded = means there are restrictions, correct?

Answer 10

I think you mean the domain of f(x)?

Answer 11

no idea, i have no sleep on me xD

Answer 12

@TuringTest Care to help with this? I'm a bit exhausted... >.<

Answer 13

Heh, it's ok. I appreciate that you try to help :)

Answer 14

just take the limit as x goes to infinity and -infinity

Answer 15

\[f(x)=2x\sin(\frac1{x^2})-{\cos(\frac1{x^2})\over x^2}-\frac12\]

Answer 16

Taking said limits would lead to an assumption that f(x) is bounded as the limits are both equal to 0

Answer 17

The area where f(x) takes values close to infinity and negative infinity is near 0

Answer 18

how is\[\lim_{x\to\infty} f(x)=0\]?

Answer 19

that's not what the question is asking, they want to know if the funciton is bounded as x ranges from -infty to +inty

Answer 20

yes

Answer 21

\[\lim_{x\to\infty}f(x)=\lim_{x\to\infty}2x\sin(\frac1{x^2})-\lim_{x\to\infty}{\cos(\frac1{x^2})\over x^2}-\lim_{x\to\infty}\frac12\]what is the value of each limit?

Answer 22

its clear from the graph that the limits are going to 0

Answer 23

yeah, except they don't

Answer 24

okay they go to -1/2 and 1/2, but how does that prove anything?

Answer 25

they don't do that either, what is\[\lim_{x\to\infty}2x\sin(\frac1{x^2})\]?

Answer 26

0?

Answer 27

right, and as \(x\to-\infty\)

Answer 28

?

Answer 29

same i guess

Answer 30

yep, and the next term?

Answer 31

to 0 as well

Answer 32

right so we are left with... ?

Answer 33

-1/2

Answer 34

for both +/- infty, yes
therefore the function is bounded both above and below by that number

Answer 35

dude, i think you are mistaken. we can only predict the range of a function using limits only if its monotonous.

Answer 36

oh I see, I did end behavior, so you want the range?

Answer 37

yeah if we manage to show that the range is a not bounded then its proven

Answer 38

then take the limit as x goes to 0, since that is where we are likely to find the maximum values of the function (when we have some 0's in the denom)

Answer 39

hmmm, the limit as x approaches 0 doesnt exist as it fluctuates between -infinity and infinity

Answer 40

does that mean anything?

Answer 41

i mean can this be used as proof?

Answer 42

yeah, it means the range is unbounded

Answer 43

yes, showing that the left hand limit and right hand limit of the cosine term approach +/- infty means that the function has no upper nor lower bound.

Answer 44

GREAT. THATS GOOD HELP :)

Answer 45

sorry I sent you off-track there for a while

Answer 46

no problem we got to the point and that helped me so much!