Question

@ganeshie8 in reference to http://openstudy.com/study#/updates/54585373e4b017f3f580c7b6 I need to find a counterexample where the limit exists, but f is not differentiable.

Answer 1

I'm having issues with the limit existing part

Answer 2

are you able to explain the situation a little?

Answer 3

\[f'(c)=\lim_{h\rightarrow 0} \frac{f(c+h)-f(c-h)}{2h}\] this is the limit in question. So my issue is I can't use |x| and I would like to use a piecewise that looks like this:
but I do not know how to define it so that it isn't diff

Answer 4

if the limit \(\lim\limits_{h\to 0}\dfrac{f(x+h)-f(x)}{h}\)exists then the function is differentiable, right ?

Answer 5

what do you mean by finding a counter example ? there are no counter examples

Answer 6

the limit is DNE at the vertex point for the graph you have drawn and at the vertex point of function |x|, so they are not differentiable at those specific points

Answer 7

may be im not interpreting ur question properly hmm

Answer 8

i think OP is mixing btw limit at a point and tangent , but in this case the limit does not refers to the limit that in the f'(c) def

Answer 9

So verbatum from the book(the reason you are confused ganeshie is the reason I am)
"b.)Show by example that the above limit may exist even if f is not differentiable at c."

Answer 10

I cannot think of how the limit can exist other than a piecewise, but if you define the point, then it's sort of differentiable

Answer 11

can u use ceiling and floor functions ? ( but for those limit does not exist ) xD

Answer 12

hmmm sin 1/x ?

Answer 13

hmm maybe, limit will exist there

Answer 14

i still believe the existence of limit of difference quotient as a neccessary and sufficient condition for differentiability for a function in one variable. but i could be wrong though

can u show me ANY example(piecewise is also fine) where the limit exists but the function is not differentiable ?

Answer 15

try on 0

Answer 16

but it isn't diff at 0. Brilliant

Answer 17

yeah , im not sure about limit
since we only say not diff but we dont use limit test

Answer 18

And ganesh, I'm honestly quite confused by it. I am going to do the calculations for sin(1/x) but my only concern is the 1/0 issue on the limit

Answer 19

u might need this identity
limit (sin x )/x=1

Answer 20

i guess xD

Answer 21

hmm, I'm still getting lim sin 1/0

Answer 22

which dne correct?

Answer 23

sin(1/x) is not continuous at x = 0.
x = 0 is not in the domain.

Answer 24

Ok, let's rethink this. What ways can a function not be differentiable?
1. discontinuity
2. corner
3. abrupt change of slope

Answer 25

any others?

Answer 26

are we still talking about the limit of difference quotient
or the limit of a function ?

Answer 27

the existence of limit of a function tells you the function is continuous

Answer 28

the existence of limit of difference quotient tells you the funciton is differentiable
(you cannot find any counter examples to this)

Answer 29

That's what I thought, but obviously there are cases where it isn't because that is the requirement for the problem

Answer 30

\( \large {\lim_{h\rightarrow 0} \frac{f( h)-f(-h)}{2h} }\)=\( \large {\lim_{h\rightarrow 0} \frac{f( h)+f(h)}{2h} }\) \\ =\( \large {\lim_{h\rightarrow 0} 2\frac{f( h) }{2h} }= \large {\lim_{h\rightarrow 0} \frac{f( h) }{ h} }\)
hmmm

Answer 31

let me get this straight :)

claim : existence of \(\lim\limits_{h\to 0}\dfrac{f(x+h)-f(x)}{h} \) need not imply differentiability

is that right ?

Answer 32

claim : f(x) is not deferentiable at C but limit of def exist xD

Answer 33

slightly different but derived from that \[f′(c)=lim_{h→0}\frac{f(c+h)−f(c−h)}{2h}\]

Answer 34

Ohk, we're messing with this definition

Answer 35

gr.. that is evil. Let me retry: \[f'(c)=\lim_{h\rightarrow 0}\frac{f(c+h)-f(c-h)}{2h}\]

Answer 36

Ok, you have said earlier that |x| function is a counter example, could u elaborate on that ?

Answer 37

I thought it was, but the limit still DNE

Answer 38

okay, i understand what we are after finaly, thnks :)

Answer 39

what about saying any smooth function that is undefined at point c , its not deferentiable at c , but limit exist xD ?

Answer 40

it's cool, sorry for the awful description

Answer 41

if it is not in the domain, why bother about differentiability ?

Answer 42

I need a specific. But the limit would exist in that case wouldn't it?

Answer 43

Doesn't that imply differntiability?

Answer 44

Oh wait!

Answer 45

you're looking for a point in the domain of function where the limit exists but the derivative doesn't

Answer 46

when c is not it domain then function is not defrentiable at that point :D
bu limit test work

Answer 47

I was thinking what ganeshie was. I think that will work! Let' me try one

Answer 48

that should work i thikn ?

Answer 49

I was trying \[f(x)={x^2~~~~x\not =0;3~~~x=0} \]

Answer 50

What about this piecewise function:

f(x) = 5 for x < 5
f(x) = x for x > 5

f(c+h) - f(c-h) = f(5+h) - f(5-h) = 5+h - 5 = h
h/2h = 1/2

Answer 51

it work but idk if that what the question want

Answer 52

@aum, but the function isn't defined at 5? Can we use that?

Answer 53

I wanted to use a piecewise, but I'm just not sure of the rules for the set up of the equation

Answer 54

for aum's example, the limit wont exist because
left limit = 0
right limit = 1

Answer 55

But the function is continuous at x = 5.
If you want you can change the first or the second definition to include x = 5.