Question

f(x)=x^(2)*(sin(1/x)) is it differentiable at x=0

Answer 1

a function being differentiable at a given point means the derivative is continuous at that point. the derivative of f is 2xsin(1/x)-cos(1/x). it's fairly straightforward to check continuity at 0.

Answer 2

thanks cnknd!! but i have doubt that does cos(1/x) also wobbles between 0 ,-1 and 1 like sin1/x

Answer 3

ok, lets break that thing into 2 parts
first: 2xsin(1/x), is this continuous at 0?

Answer 4

yes

Answer 5

2nd part, cos(1/x): is this continuous at 0?

Answer 6

remember what it means to be continuous: the left and right limits exist and are equal

Answer 7

cos1/x is similar to sin1/x as x approaches to 0 i.e it keeps oscillating between 0 ,1 ,-1 as x approaches 0

Answer 8

well the answer here is that neither the left nor right limits exist for cos(1/x) as x goes to 0
(showing why this is true is a bit more involved, let me know if u want to go through the rigor)

Answer 9

yes please that will be helpful

Answer 10

ok, so let's look at the right limit of cos(1/x) as x -> 0
i'll first make a change of variables: let y = 1/x

Answer 11

so this limit now becomes: limit of cos(y) as y -> positive infinity

Answer 12

well, i havent specified how y is going to infinity. perhaps it goes as the following sequence: 1, 2, 3, 4, ....; or this sequence: 3, 5, 7, 9, 11, ...

Answer 13

if i can identify 2 ways (2 different sequences) for y to approach infinity, but the 2 paths would give different limits of the function cos(y), then it doesn't make sense to say that the limit of the function exists

Answer 14

following me so far?

Answer 15

i dint get that 2 sequences

Answer 16

a sequence here just means how y is going to infinity.

Answer 17

so here's one way for y to approach infinity:
y = 0, 2pi, 4pi, 6pi, .....
and so the value of cos(y), as y approaches infinity in this path, goes as:
cos(y) = 1, 1, 1, 1, ...
so without going further, have i convinced you that the limit of cos(y) is 1?

Answer 18

yes i got that

Answer 19

well here's another sequence for y:
y = pi, 3pi, 5pi, ....
so we have:
cos(y) = -1, -1, -1, ...
and it looks like the limit here (for the same function) is -1.

the limit can't possibly be -1 and 1 at the same time. therefore it's undefined

Answer 20

y dint u take npi/2 for various values of n in consideration?

Answer 21

i could, and that would give me a limit of 0

Answer 22

still the same conclusion though, the limit can't be -1, 0, and 1 at the same time, so it does not exist

Answer 23

ok thanks cnknd!!! that means f(x) is not differentiable at x= 0

Answer 24

that is correct.

Answer 25

in one of the class notes i saw that
lhl, f'(0) =limf(0-h)-f(0)/h=h*sin(1/h) =0 ;h --> 0

Answer 26

and rhl, f'(0)=lim f(0+h) - f(0)/h =h* sin1/h = 0

Answer 27

lhl= rhl thereby showing that f'(x)is continuous at x= 0 , i cant find the mistake

Answer 28

i have no idea how they did this step: (f(0+h)-f(0))/h = h*sin(1/h)

Answer 29

they have shown f(0+h)= h^2 * sin(1/h) aand f(0)= 0 and then divided by h

Answer 30

o.o sorry i think i was wrong...
the wikipedia page has this function as an example of differentiable but not smooth:
http://en.wikipedia.org/wiki/Differentiable_function

i apologize for misleading you

Answer 31

but even in your approach i didnt find anything wrong!!
i am really confused

Answer 32

oh i went wrong in the definition: i thought it was asking for smooth (i.e. differentiable, and derivative is continuous). where as it was simply asking for differentiability

Answer 33

being differentiable and derivative is continuous?
are they both different?

Answer 34

the distinction here is not a place in mathematics that I've looked at in detail. but wikipedia says a function can have certain kinds of discontinuity in the derivative, and still be differentiable:
http://en.wikipedia.org/wiki/Differentiable_function#Differentiability_classes

Answer 35

so to test differentiability what should be my approach for any given f(x)

Answer 36

ur lecture notes would be correct for simple differentiability
i.e. use the definition of the derivative.

Answer 37

i still dont understand that if we take derivative of f(x) and try to check its continuity it is discontinuous....and yet it is continuous using the defintion of derivative
why two different answer?

Answer 38

oh the definition of the derivative doesnt say that the derivative is continuous

Answer 39

the limits in the definition of the derivative is telling you whether or not your approximations of the derivative is continuous (as they get more and more accurate to the true derivative)

Answer 40

but y two different answer...because even the former approach seems to be correct

Answer 41

well it's not asking for whether or not the derivative is continuous, it's just asking for whether or not the derivative exists

Answer 42

ok to test differentiability i should use limf(x+h)-f(h)/h as h --> 0

Answer 43

yes

Answer 44

ok thanks cnknd!!