# Thread: Derivatives, differentiabilty, continuity, boundedness

1. ## Derivatives, differentiabilty, continuity, boundedness

Let g(x) =x^asin(1/x) if x is not 0
g(x)=0 if x=0

Find a particular value for a such that
a) g is differentiable on R but such that g' is unbounded on [0,1].
b) g is differentiable on R with g' continuous but not differentiable at zero
c) g is differentiable on R but and g' is differentiable on R, but such that g'' is not continuous at zero.

I've only started on a) but I got a little stuck
Well I know that a function is differentiable if g' exists for all points.
I started by calculating derivatives:
g'(x)=ax^(a-1)sin(1/x)-x^(a-2)cos(1/x) if x is not zero.
Now for x=0, we have:
g'(0)=limg(x)-g(0)/x-0=limg(x)/x=limx^(a-1)sin(1/x)
I know limxsin(1/x) does not exist, but for all other x, g'(0) does exist. Then g'(0) exists and equals 0 provided a>2.
So g is differentiable provided a>2.
Now the part that confuses me is the unboundedness. I looked in this particular chapter of my book and there was no mention of boundedness at all.
So from previous chapters, I had that a sequence (xn) is bounded if there exists a number M>0 such that |xn|<=M for all n in N. I just am really confused on the unboundedness part.

2. Originally Posted by mathematic
Let g(x) =x^asin(1/x) if x is not 0
g(x)=0 if x=0

Find a particular value for a such that
a) g is differentiable on R but such that g' is unbounded on [0,1].
b) g is differentiable on R with g' continuous but not differentiable at zero
c) g is differentiable on R but and g' is differentiable on R, but such that g'' is not continuous at zero.

I've only started on a) but I got a little stuck
Well I know that a function is differentiable if g' exists for all points.
I started by calculating derivatives:
g'(x)=ax^(a-1)sin(1/x)-x^(a-2)cos(1/x) if x is not zero.
Now for x=0, we have:
g'(0)=limg(x)-g(0)/x-0=limg(x)/x=limx^(a-1)sin(1/x)
I know limxsin(1/x) does not exist, but for all other x, g'(0) does exist. Then g'(0) exists and equals 0 provided a>2.
So g is differentiable provided a>2.
Now the part that confuses me is the unboundedness. I looked in this particular chapter of my book and there was no mention of boundedness at all.
So from previous chapters, I had that a sequence (xn) is bounded if there exists a number M>0 such that |xn|<=M for all n in N. I just am really confused on the unboundedness part.
Since $g(x)$ is continuous and smooth except at $x=0$ it is differentiable on $x \in\{ \mathbb{R}\setminus 0 \}$

Now at $x=0$
$\displaystyle g'(x)=\lim_{x \to 0}\frac{g(x)-g(0)}{x-0}=\lim_{x\to 0}\frac{x^a\sin\left( \frac{1}{x}\right)}{x}=\lim_{x \to 0}x^{a-1}\sin\left( \frac{1}{x}\right)$

This gives

$\displaystyle\lim_{x \to 0} -x^{a-1}\le \lim_{x \to 0}x^{a-1}\sin\left( \frac{1}{x}\right) \le \lim_{x \to 0}x^{a-1}$

This limit goes to zero if $a>1$. How does this help with part a)?

3. Originally Posted by mathematic
I know limxsin(1/x) does not exist, but...
$\displaystyle \lim_{x\to 0} x\sin(1/x)=0$

In fact even more could be said about similar limits, but I see that Mr Set has already covered that.

CB

4. Originally Posted by TheEmptySet
This gives

$\displaystyle\lim_{x \to 0} -x^{a-1}\le \lim_{x \to 0}x^{a-1}\sin\left( \frac{1}{x}\right) \le \lim_{x \to 0}x^{a-1}$

This limit goes to zero if $a>1$. How does this help with part a)?
I don't really follow how you got that. But from that, I'm thinking a>1.

5. Originally Posted by CaptainBlack
$\displaystyle \lim_{x\to 0} x\sin(1/x)=0$

In fact even more could be said about similar limits, but I see that Mr Set has already covered that.

CB
Oh I see where I went wrong with that statement. I was thinking if we had f(x)=xsin(1/x) and f(0)=0, then f'(0) DNE.

6. g' unbounded means we want for M>0, there exists x0 in [0,1] such that |g'(x0)|>M