Differentiability and continuity confusion

[physics4life]

Hi there just a general question: this involves continuity and differentiability

suppose:

f(x) = sin 1/x if x not equal to 0
f(0) = 0

PROVE F IS NOT DIFFERENTIABLE AT 0


i understand if it is not differentiable at 0 then it may not be continuous at 0. however is there a way of probing f is not diff at 0 and seeing whether it is continuous at 0 or not?
thnx a lot

 

[cristo]

What is the definition of a derivative at a point?

 

[physics4life]

a function is differentiable if [ f(x) - f(xo) / (x - x0) ]


exists
i can appreciate i have to do this.. but how would i show that this derivative doesn't exist at 0... n wuld it be continuous at that point or not?but from here

 

[Kummer]

You need to show,
$$\lim_{x\to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x\to 0} \frac{\sin \frac{1}{x}}{x}$$
Does not exist.

Note: I think your function is $$f(x) = x\sin \frac{1}{x}$$. Because this is a standard example.

 

[VietDao29]

You need to show,
$$\lim_{x\to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x\to 0} \frac{\sin \frac{1}{x}}{x}$$
Does not exist.

Note: I think your function is $$f(x) = x\sin \frac{1}{x}$$. Because this is a standard example.

Yeah, the function f(x) = sin(1/x) is even discontinuous at 0, and hence, not differentiable at x = 0. Are you sure your function isn't f(x) = x sin(1/x)?

 

[CompuChip]

i understand if it is not differentiable at 0 then it may not be continuous at 0.

Be careful here.

Any differentiable function is continuous. The negation is: if it's not continuous then it's not differentiable -- this is a consequence of the definition: if f is differentiable at a then
$$\lim_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(a)$$
exists so in particualar
$$\lim_{x \to a} f(x) = \lim_{x \to a} f(a) + f'(a) (x - a) = f(a)$$.

The converse however, (if it's not differentiable then it's not continuous) is not true, for example, the absolute value function $x \mapsto |x|$ is not differentiable (at 0) but it is very continuous.

 

[physics4life]

the function is definitely sin(1/x) and NOT x.sin(1/x)

so basically am i right in assuming that to answer a question like this (show its not differentiable at 0), i have to first show that its discontinuous at 0 and deduce that since its discontinuous at 0 then it can't be differentiable at 0? is that the right path to answer a question like this or do i have to take another path?

but how did Viet deduce it's even discontinuous at 0.. I am assuming you took the left hand limit and the right hand limits and they didn't equal the same? hence the deduction that it is discontinous? which then deduces that the function is not differentiable at 0 (going by compuchip where it was said "The negation is: if it's not continuous then it's not differentiable")please enlighten me lol. i think I am starting to get the hang of this, just need to get over the mental block.

 

[CompuChip]

Personally I would use the definition of limits.
Suppose it is continuous. Then sin(1/x) goes to 0 as x goes to 0 would mean the following: for any $\epsilon$ > 0 there is a $\delta$ > 0 such that $|x| < \delta$ implies $|f(x)| < \epsilon$.

So let's prove it's not continuous (nice practice in negating statements like the one above ): let $0 < \epsilon < 1$. Then for any $\delta > 0$ there is an $x, |x| < \delta$ such that $|f(x)| > \epsilon$.

By the way, note that you don't have to show it's not continuous to show it's not differentiable. It's just one way of doing it.

 

[pardesi]

the function is not differentialble because it is not continious that's because
$$\lim _{x \to \infty sin x$$ doesn't exist .that's because lets' say x tends to $$\infty$$ by assuming values of the form $$2n \pi$$ then the limit is 1 but if i change that to $$\frac{(2n+1) \pi}{2}$$ then it becomes 0.so the limit doen't exist so the function is not continious there

 

[physics4life]

thnx for all the advice.. I've done some research on this and I've just still got a bit of problem here...

i want to prove that sin(1/x) is not differentiable by using the LIMITS method..

ive worked this out so far:

the left hand limit and right hand limits need to be proved different so that we can conclude that sin(1/x) is not differentiable at 0

but my problem is i keep getting the same limits for the left and right

limit from the right:

$$\lim_{h\to 0+} \frac{f(0+h)-f(0)}{h} = \lim_{h\to 0+} \frac{{\sin \frac{1}{0+h}}-f(0)}{h} = \lim_{h\to 0+} \frac{\sin(1)}{0} = \sin(1)$$

limit from the left:

$$\lim_{h\to 0-} \frac{f(0+h)-f(0)}{h} = \lim_{h\to 0-} \frac{{\sin \frac{1}{0+h}}-f(0)}{h} = \lim_{h\to 0-} \frac{\sin(1)}{0} = \sin(1)$$

-----------------------------------------------------------
so why am i getting the same limits. presumably if its not differentiable then the 2 limits have to be different?

thnx a lot for the help. if i can be cleared about what may be going wrong here then it'd clear a lot of confusion in my mind

 

[DeadWolfe]

Why on Earth would you say sin(1)/0=sin(1)?

Apparently 0=1?

 

[d_leet]

Why on Earth would you say sin(1)/0=sin(1)?

Apparently 0=1?

Moreso why would you say that sin(1/(0+h)) is equal to sin(1)?

 

[physics4life]

i was attempting to cancel everything out at the end without considering obvious facts in the beginning but i guess that would be a mistake..

for the right hand limit:
$$\lim_{h\to 0+} \frac{f(0+h)-f(0)}{h} = \lim_{h\to 0+} \frac{{\sin \frac{1}{0+h}}-f(0)}{h} = \lim_{h\to 0+} \frac{{\sin \frac{1}{h}}}{h}$$

from here since h tends to 0... i have [sin(1/0) / 0]..

even for the left hand limit i get the same answer...

so I am basically getting the left and right hand limit to be the same which concludes to the function being differentiable at 0. but obviously that can't be true as it contradicts the fact that the function is not differentiable at 0. so what should i acutally have done please?

 

[CompuChip]

What do you know about Taylor expansions?

 

[pardesi]

even for the left hand limit i get the same answer...

so I am basically getting the left and right hand limit to be the same which concludes to the function being differentiable at 0. but obviously that can't be true as it contradicts the fact that the function is not differentiable at 0. so what should i acutally have done please?

there is no meaning in saying
$$\inf= \inf$$
or diverging is equal to diverging
well "=" this symbol holds true just for real numbers and in case of equivalent statements but not for diverging or non existent quantities the limit as u got as i had proved earlier doesnot exist as such there is no question of an derivative existing at that point no matter it's left or right

 

[C0nfused]

i was attempting to cancel everything out at the end without considering obvious facts in the beginning but i guess that would be a mistake..

for the right hand limit:
$$\lim_{h\to 0+} \frac{f(0+h)-f(0)}{h} = \lim_{h\to 0+} \frac{{\sin \frac{1}{0+h}}-f(0)}{h} = \lim_{h\to 0+} \frac{{\sin \frac{1}{h}}}{h}$$

from here since h tends to 0... i have [sin(1/0) / 0]..

even for the left hand limit i get the same answer...

so I am basically getting the left and right hand limit to be the same which concludes to the function being differentiable at 0. but obviously that can't be true as it contradicts the fact that the function is not differentiable at 0. so what should i acutally have done please?

As pardesi mentioned, you can't say that two functions have the same limit, if this limit doesn't exist. Anyway, the thing that causes you problems is the limit of sin(1/x) when x->0.

As pardesi also mentioned, this limit doesn't exist. There's a theroem that says that a limit of a function f(x) exists at x0 iff for any sequence a(n), a(n)<>x0 , n=0,1,... with lima(n)=x0 , n->infinity, the limit of f(a(n)) exists and limf(a(n))=x0, n->infinity.

So if you consider a(n)=1/2πn and b(n)=1/(2n+1/2)π then lima(n)=0=limb(n) and f(a(n))=sin2πn=0 and f(b(n))=sin(2πn+π)=1 so limf(a(n))=0 and limf(b(n))=1. Thus, the limit at x0=0 doesn't exist.

After you get this, you have to consider the limit of g(x)=sin(1/x)/x,x->0. Considering the same two sequences we get g(a(n))=0 and g(b(n))=(2n+1/2)π, so limg(a(n))=0 and limg(b(n))=+infinity. So the limit doesn't exist, so the function sin(1/x) is not differentiable at x=0

 

[mathusers]

As pardesi also mentioned, this limit doesn't exist. There's a theroem that says that a limit of a function f(x) exists at x0 iff for any sequence a(n), a(n)<>x0 , n=0,1,... with lima(n)=x0 , n->infinity, the limit of f(a(n)) exists and limf(a(n))=x0, n->infinity.

C0nfused.. what is this theorem called please? so that I can research this aswell. I'm having a bit of a problem on differentiability too so this thread will definitely help my understanding. Also I can't seem to process your explanation of this theorem you were explaining, please forgive. Maybe if you can put that in latex form then i can see what you were trying to show?
Cheers :)

 

[Extropy]

Actually, there is no standard name for this theorem.

Now, you should already be aware of the theorem that states:

$$\forall$$F (F is a function -> $$\forall$$x (F is differentiable at x -> F is continuous at x))[/B]

Since you already know that Sin($$\frac{1}{x}$$) is a function, you need only to prove that $$\neg$$"F is continuous at x" since "$$\neg$$(F is continuous at x) -> $$\neg$$(F is differentiable at x)" is the contrapositive of and is thus logically equivalent to "F is differentiable at x -> F is continuous at x."

The definition of continuity at a point $$x_0$$ is "$$\forall \epsilon>0 \exists \delta>0 \forall x (|x-x_0|<\delta \rightarrow |f(x)-f(x_0)|<\epsilon)$$

Therefore, you need to prove that $$\exists \epsilon>0\forall\delta>0\exists x (|x-x_0|<\delta\wedge|f(x)-f(x_0)|\ge\epsilon).$$ This is not too difficult, since $$\epsilon<\frac{1}{2}$$ will suffice. Now, you only need to prove that this works.

 

[Feldoh]

As pardesi mentioned, you can't say that two functions have the same limit, if this limit doesn't exist. Anyway, the thing that causes you problems is the limit of sin(1/x) when x->0.

As pardesi also mentioned, this limit doesn't exist. There's a theroem that says that a limit of a function f(x) exists at x0 iff for any sequence a(n), a(n)<>x0 , n=0,1,... with lima(n)=x0 , n->infinity, the limit of f(a(n)) exists and limf(a(n))=x0, n->infinity.

So if you consider a(n)=1/2πn and b(n)=1/(2n+1/2)π then lima(n)=0=limb(n) and f(a(n))=sin2πn=0 and f(b(n))=sin(2πn+π)=1 so limf(a(n))=0 and limf(b(n))=1. Thus, the limit at x0=0 doesn't exist.

After you get this, you have to consider the limit of g(x)=sin(1/x)/x,x->0. Considering the same two sequences we get g(a(n))=0 and g(b(n))=(2n+1/2)π, so limg(a(n))=0 and limg(b(n))=+infinity. So the limit doesn't exist, so the function sin(1/x) is not differentiable at x=0

For clarity and simplicity:
$$\lim_{x\to c}f(x) = L$$ exists iff $$\lim_{x\to c^+}f(x) = L$$ and $$\lim_{x\to c^-}f(x) = L$$

 

[HallsofIvy]

As pardesi mentioned, you can't say that two functions have the same limit, if this limit doesn't exist. Anyway, the thing that causes you problems is the limit of sin(1/x) when x->0.

As pardesi also mentioned, this limit doesn't exist. There's a theroem that says that a limit of a function f(x) exists at x0 iff for any sequence a(n), a(n)<>x0 , n=0,1,... with lima(n)=x0 , n->infinity, the limit of f(a(n)) exists and limf(a(n))=x0, n->infinity.

So if you consider a(n)=1/2πn and b(n)=1/(2n+1/2)π then lima(n)=0=limb(n) and f(a(n))=sin2πn=0 and f(b(n))=sin(2πn+π)=1 so limf(a(n))=0 and limf(b(n))=1. Thus, the limit at x0=0 doesn't exist.

After you get this, you have to consider the limit of g(x)=sin(1/x)/x,x->0. Considering the same two sequences we get g(a(n))=0 and g(b(n))=(2n+1/2)π, so limg(a(n))=0 and limg(b(n))=+infinity. So the limit doesn't exist, so the function sin(1/x) is not differentiable at x=0

For clarity and simplicity:
$$\lim_{x\to c}f(x) = L$$ exists iff $$\lim_{x\to c^+}f(x) = L$$ and $$\lim_{x\to c^-}f(x) = L$$

No, that not what the first quote said. What the first quote said was:
$$\lim_{x|to c}f(x)$$ exists, and is equal to L, iff for every sequence {x_n} converging to c, f(x_n) converges to L.

It said nothing about one sided limits.