Question

Limit sin(1/x) as x->0 diverges, right? so is it continuous where x->0?

Answer 1

continuous implies something about the value at x=0

Answer 2

its continuous as it approaches zero, yeah .... it just alternates between -1 and 1 and never settles between the two

Answer 3

Thanks. I think there's a mistake in the question then. It says that it is continuous where x=0, but when x=0 then its undefined, right?

Answer 4

1/x is undefined at x=0 and as such, there is no value to attribute to the sine of an undefinable angle

Answer 5

the limit of a function doesnt depend on the value of a function at x, it depends on if you can get the same value by aproaching it from all directions

Answer 6

is the function continuous as it approaches 1/0? sure, for sin is defined for all values that are definable

Answer 7

the question is actually this one: Prove that the function f(x)=x(sin(1/x), when x is not 0, f(x)=0 when x=0, is continuous at x=0 but isnt differentiable at the same point

Answer 8

sorry for poor translation

Answer 9

so we seem to have a squeeze thrm here

Answer 10

the amplitude of the function goes to zero, so the sin function will fluctuate between the lines y=x and y=-x which of course funnels in at x=0

Answer 11

differentiable, id assume we just take a derivative and see what it results

Answer 12

not sure what methods are availble to you

Answer 13

In this case I was told to use the definition of the derivative. It wasn't differentiable

Answer 14

sin(u) = u - u^3/3 + u^5/5 - u^7/7 + ...

sin(1/x) = 1/x - 1/3x^3 + 1/5x^5 - 1/7x^7 + ...

xsin(1/x) = 1 - 1/3x^2 + 1/5x^4 - 1/7x^6 + ...

that would be the taylor polynomial for it if that helps lol

Answer 15

Yes, it does. But the tutor told me to use the definition f'(x)=lim (f(x+h) - f(x))/h as h->0
and doing that i end up with lim sin(1/x) as x->0

Answer 16

i forgot those were factorials ... not digits

Answer 17

Dont worry

Answer 18

x sin(1/x)

x' sin(1/x) + xcos(1/x) (-1/x^2)

sin(1/x) - [cos(1/x)] / x

the derivative would get us that

Answer 19

which is not definable at x=0

Answer 20

so it is not continuous at x=0?

Answer 21

are you asking if \[\sin\left(\frac{1}{x}\right)\] is continuous at 0?

Answer 22

you keep asking for continuous, while speaking of limits

Answer 23

if the limit of f(x), as x approaches c+ and c-, is equal to f(c) then it is continuous at x=c if memory serves for a definition of continuity :)

Answer 24

The question asks to prove that the function f(x)= x(sin(1/x)) for x is not 0, and f(x)=0 for x=0, is continuous but not differentiable. so its assuming its continuous

Answer 25

continuous at x=0 but not diff. at the same point

Answer 26

It's the fourth question. translation "prove that the function f(x) is continuous at x=0 but not differentiable at the same point"

Answer 27

ah, you have a piecewise function

Answer 28

by the squeeze thrm, the limit of f(x) as x approaches 0 from the left and right is ... 0

and by definition of f(x), f(0)=0

since the limit and the function are the same at x=0, then it is continuous at x=0 by the crude recollection of a definition of continuity

Answer 29

hmm that makes sense now. but it's also not differentiable, right?

Answer 30

at x=0

Answer 31

derivatives deal with slopes, can we conclude what the slope of the line is at x=0?

that is where the limit definition would prolly be applied

Answer 32

its like trying to find the derivative of |x| at x=0

Answer 33

at any rate ... sin(1/x) - [cos(1/x)]/x is our derivative

and successive derivatives are just as absurd.

the taylor polynomial helps to see this better

Answer 34

helps ME to see this better that is :)

Answer 35

just apply the definition of the derivative

you will end up with computing the limit\[\lim_{h\to 0}\sin\left(\frac{1}{h}\right)\]

which is easy to do

Answer 36

I see. Look, this is the answer they told me is the right one:
hmm. lim (h -> 0) f(h) / h = lim (h -> 0) h*sen(1/h)/h
so
lim (h->0) sen(1/h)

1/h -> infinity when h->0

lim (x-> infinity) sen(x) does not exist, so its not differentiable at x=0

Answer 37

looks like zarkons reply :)

Answer 38

hmm makes sense now. so lim sin(infinity) does not exist?

Answer 39

the sine function if two pi periodic. so you just keep getting the same stuff over and over as x goes to infinity. it never converges

Answer 40

It diverges, right? I just don't get how it is continuous if it never converges. I'm kind of new to this.

Answer 41

can you determine the slope of |x| at x=0? continuity and limit are 2 different things.

Answer 42

as well as slope

Answer 43

hmm. I understand. Thanks to you all.

Answer 44

yw