- #1
bluecode
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Homework Statement
Refer to attached file.
The attempt at a solution
(a)
[tex]
g'(0) = \lim_{x\rightarrow 0} {\frac{g(x)-g(0)}{x-0}}
[/tex]
[tex]
g'(0) = \lim_{x\rightarrow 0} {\frac{x^\alpha cos(1/x^2)-0}{x}}
[/tex]
[tex]
g'(0) = \lim_{x\rightarrow 0} {x^{(\alpha-1)} cos(1/x^2)-0}
[/tex]
[tex]
g'(0) = 0
[/tex]
So I've attempted to show that [itex]g'(0) = 0[/itex] and therefore that [itex]g'(x)[/itex] is differentiable at [itex]x = 0[/itex]. In this case, would the values of [itex]\alpha[/itex] for which this condition holds simply be [itex]\alpha> 0[/itex]? It seems to simple for this to be the case.
(b)
My interpretation of "continuously differentiable" is that at [itex]x = 0[/itex], [itex]g(x)[/itex] must be continuous and [itex]g'(x)[/itex] must exist. The latter condition is implied from part (a). So I just need to prove that [itex]g'(x)[/itex] is continuous at [itex]x = 0[/itex].
[tex]
-1 ≤ cos(1/x^2) ≤ 1
[/tex]
[tex]
-x^\alpha ≤ x^{\alpha}cos(1/x^2) ≤ x^\alpha
[/tex]
[tex]
\lim_{x\rightarrow 0} {-x^\alpha} = 0
[/tex]
and
[tex]
\lim_{x\rightarrow 0} {x^\alpha} = 0
[/tex]
so by the Squeeze Law,
[tex]
\lim_{x\rightarrow 0} {x^\alpha cos(1/x^2)} = 0 = g(0)
[/tex]
For this part, it's a similar scenario.
I've proven that [itex]g'(x)[/itex] is continuously differentiable at [itex]x = 0[/itex], but for this to happen, [itex]\alpha[/itex] can take any value greater than 0, can't it?
I'd be grateful for any assistance with this question. Thanks in advance!
Refer to attached file.
The attempt at a solution
(a)
[tex]
g'(0) = \lim_{x\rightarrow 0} {\frac{g(x)-g(0)}{x-0}}
[/tex]
[tex]
g'(0) = \lim_{x\rightarrow 0} {\frac{x^\alpha cos(1/x^2)-0}{x}}
[/tex]
[tex]
g'(0) = \lim_{x\rightarrow 0} {x^{(\alpha-1)} cos(1/x^2)-0}
[/tex]
[tex]
g'(0) = 0
[/tex]
So I've attempted to show that [itex]g'(0) = 0[/itex] and therefore that [itex]g'(x)[/itex] is differentiable at [itex]x = 0[/itex]. In this case, would the values of [itex]\alpha[/itex] for which this condition holds simply be [itex]\alpha> 0[/itex]? It seems to simple for this to be the case.
(b)
My interpretation of "continuously differentiable" is that at [itex]x = 0[/itex], [itex]g(x)[/itex] must be continuous and [itex]g'(x)[/itex] must exist. The latter condition is implied from part (a). So I just need to prove that [itex]g'(x)[/itex] is continuous at [itex]x = 0[/itex].
[tex]
-1 ≤ cos(1/x^2) ≤ 1
[/tex]
[tex]
-x^\alpha ≤ x^{\alpha}cos(1/x^2) ≤ x^\alpha
[/tex]
[tex]
\lim_{x\rightarrow 0} {-x^\alpha} = 0
[/tex]
and
[tex]
\lim_{x\rightarrow 0} {x^\alpha} = 0
[/tex]
so by the Squeeze Law,
[tex]
\lim_{x\rightarrow 0} {x^\alpha cos(1/x^2)} = 0 = g(0)
[/tex]
For this part, it's a similar scenario.
I've proven that [itex]g'(x)[/itex] is continuously differentiable at [itex]x = 0[/itex], but for this to happen, [itex]\alpha[/itex] can take any value greater than 0, can't it?
I'd be grateful for any assistance with this question. Thanks in advance!