Differentiability and Continuity at a point

[bluecode]

Homework Statement

Refer to attached file.

The attempt at a solution
(a)
<br /> g&#039;(0) = \lim_{x\rightarrow 0} {\frac{g(x)-g(0)}{x-0}}<br />
<br /> g&#039;(0) = \lim_{x\rightarrow 0} {\frac{x^\alpha cos(1/x^2)-0}{x}}<br />
<br /> g&#039;(0) = \lim_{x\rightarrow 0} {x^{(\alpha-1)} cos(1/x^2)-0}<br />
<br /> g&#039;(0) = 0<br />
So I've attempted to show that g&#039;(0) = 0 and therefore that g&#039;(x) is differentiable at x = 0. In this case, would the values of \alpha for which this condition holds simply be \alpha&gt; 0? It seems to simple for this to be the case.

(b)
My interpretation of "continuously differentiable" is that at x = 0, g(x) must be continuous and g&#039;(x) must exist. The latter condition is implied from part (a). So I just need to prove that g&#039;(x) is continuous at x = 0.
<br /> -1 ≤ cos(1/x^2) ≤ 1<br />
<br /> -x^\alpha ≤ x^{\alpha}cos(1/x^2) ≤ x^\alpha<br />
<br /> \lim_{x\rightarrow 0} {-x^\alpha} = 0 <br />
and
<br /> \lim_{x\rightarrow 0} {x^\alpha} = 0 <br />
so by the Squeeze Law,
<br /> \lim_{x\rightarrow 0} {x^\alpha cos(1/x^2)} = 0 = g(0)<br />
For this part, it's a similar scenario.

I've proven that g&#039;(x) is continuously differentiable at x = 0, but for this to happen, \alpha can take any value greater than 0, can't it?

I'd be grateful for any assistance with this question. Thanks in advance!

 

[hunt_mat]

What if 0&lt;\alpha\leqslant 1? Then the limit wouldn't exist, why? Take a look when would happen if \alpha =1 for example.

 

[HallsofIvy]

Homework Statement

Refer to attached file.

The attempt at a solution
(a)
<br /> g&#039;(0) = \lim_{x\rightarrow 0} {\frac{g(x)-g(0)}{x-0}}<br />
<br /> g&#039;(0) = \lim_{x\rightarrow 0} {\frac{x^\alpha cos(1/x^2)-0}{x}}<br />
<br /> g&#039;(0) = \lim_{x\rightarrow 0} {x^{(\alpha-1)} cos(1/x^2)-0}<br />
<br /> g&#039;(0) = 0<br />
So I've attempted to show that g&#039;(0) = 0 and therefore that g&#039;(x) is differentiable at x = 0. In this case, would the values of \alpha for which this condition holds simply be \alpha&gt; 0? It seems to simple for this to be the case.

(b)
My interpretation of "continuously differentiable" is that at x = 0, g(x) must be continuous and g&#039;(x) must exist.

That is an incorrect interpretation. As you say below, any differentiable function must be continuous so, with that interpretation, "continuously differentiable" would be no different from "differentiable". "Continuously differentiable" at x= a means that f'(x) exists in some neighborhood of a and f'(x) is continuous at x= a.

The latter condition is implied from part (a). So I just need to prove that g&#039;(x) is continuous at x = 0.
<br /> -1 ≤ cos(1/x^2) ≤ 1<br />
<br /> -x^\alpha ≤ x^{\alpha}cos(1/x^2) ≤ x^\alpha<br />
<br /> \lim_{x\rightarrow 0} {-x^\alpha} = 0 <br />
and
<br /> \lim_{x\rightarrow 0} {x^\alpha} = 0 <br />
so by the Squeeze Law,
<br /> \lim_{x\rightarrow 0} {x^\alpha cos(1/x^2)} = 0 = g(0)<br />
For this part, it's a similar scenario.

I've proven that g&#039;(x) is continuously differentiable at x = 0, but for this to happen, \alpha can take any value greater than 0, can't it?

I'd be grateful for any assistance with this question. Thanks in advance!

 

[bluecode]

Thanks for the help everyone! I think I understand the question better now.

So taking another go (can someone check whether this would be correct this time?):
(a) For <br /> \lim_{x\rightarrow 0} {x^{(\alpha-1)} cos(1/x^2)}=g&#039;(0) = 0<br /> to always hold, the following must be satisfied:
<br /> -x^{(\alpha-1)} ≤ x^{(\alpha-1)}cos(1/x^2) ≤ x^{(\alpha-1)}<br />
For this to happen, \alpha &gt; 1 otherwise \lim_{x\rightarrow 0} {x^{(\alpha-1)} cos(1/x^2)} will be in between negative and positive infinity and so the limit would not exist.

(b) I understand that for g(x) to be continuously differentiable at x=0, g(x) must be differentiable at x=0 and g'(x) must be continuous at x=0.
After differentiating g(x) I eventually get:
<br /> \lim_{x\rightarrow 0} {g&#039;(x)}={\lim_{x\rightarrow 0} [{(\alpha)x^{(\alpha-1)} cos(1/x^2)} + 2x^{(\alpha-3)} sin(1/x^2)}]=0<br />
So by the same reasoning as part (a), \alpha &gt; 3

One question though: would it be OK to just say that the power of x in the inequality <br /> -x^{(\alpha-1)} ≤ x^{(\alpha-1)}cos(1/x^2) ≤ x^{(\alpha-1)}<br /> needs to be greater than 0, and so \alpha &gt; 1?