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Discontinuous Derivative
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The function $$f(x) = \begin{cases} x^2\sin{\frac{1}{x}} &&x \ne 0 \\ 0 && x = 0\end{cases}$$
is continuous everywhere because $$ \begin{align*} -x^2 \le x^2\sin{\frac{1}{x}} &\le x^2 && \forall x \\ \Rightarrow \lim_{x \rightarrow 0}{-x^2} \le \lim_{x \rightarrow 0}{x^2\sin{\frac{1}{x}}} &\le \lim_{x \rightarrow 0}{x^2} \\ \Rightarrow 0 \le \lim_{x \rightarrow 0}{x^2\sin{\frac{1}{x}}} &\le 0 \\ \Rightarrow \lim_{x \rightarrow 0}{f(x)} &= f(0) \end{align*}$$

Also, for $x \ne 0$ we have $$f'(x) = 2x\sin{\frac{1}{x}}-\cos{\frac{1}{x}}$$
And at $x = 0$ we have $$ \begin{align*} f'(0) &= \lim_{h \rightarrow 0}{\frac{f(h) - f(0)}{h}} \\ &= \lim_{h \rightarrow 0}{\frac{h^2\sin{\frac{1}{h}} - 0}{h}} \\ &= \lim_{h \rightarrow 0}{h\sin{\frac{1}{h}}} \\ &= 0 &&\text{since} \\ -h &\le h\sin{\frac{1}{h}} \le h \end{align*}$$

But $\lim_{x \rightarrow 0}{2x\sin{\frac{1}{x}}-\cos{\frac{1}{x}}}$ does not exist since $\cos{\frac{1}{x}}$ attains every value between $-1 \text{ and } 1$ in the interval $(0,\epsilon)$ no matter how small we choose $\epsilon$ to be. The same applies to the interval $(-\epsilon,0)$.

Thus $f'(x)$ exists everywhere and is continuous everywhere but is discontinuous at $x = 0$.

Mean Value Theorem
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The mean value theorem states that given $f(x)$ continuous in the closed interval $(a, a + h)$ and that $f'(x)$ exists in the open interval $(a, a + h)$, then $$f(a + h) = f(a) + hf'(a + \theta h)$$ for some $0 \lt \theta \lt 1$. This can be rewritten as $$\frac{f(a+h)-f(a)}{h}=f'(a+\theta h)$$
And the limit of the left hand side as $h \rightarrow 0$ is $f'(a)$, which means that $$\lim_{h \rightarrow 0}{f'(a+\theta h)} = f'(a)$$ which implies that f'(x) is continuous at $x = a$.

The Problem
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These two results are contradictory. The second argument clearly has a flaw since the first argument is a counter-example. The question is: what is the flaw?

Mon, 13 Jan 2014 05:43 GMT