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$$f(x)= \left\{\begin{array}{ll} x-x^2 &\mbox{if $x$ is rational,}\\ x+x^2 &\mbox{if $x$ is irrational.} \end{array}\right.$$ Show that $f'(0)=1$ and yet there is no neighborhood $I$ of the point $0$ on which this function is monotonically increasing.

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To show that $f'(0)=1$, it suffices to prove that $\lim_{x\rightarrow 0}\frac{f(x)-x}x = 0$ (since $f(0) = 0$). Since $f(x)-x=\pm x^2$, it is always true that $|f(x)-x|=x^2$. Dividing by $x$ and making $x\rightarrow 0$ gives the answer.

As for the second part. Suppose such a neighborhood $I$ exists. Then let $x\in I$ be irrational. Since $\mathbb Q$ is dense in $\mathbb R$, there exists a sequence $r_n$ of rationals in $I$ such as $r_n\rightarrow x$ and $r_n > x$.

Since $r_n$ is rational, $f(r_n) = r-r_n^2$, on the other hand $f(x) = x+x^2$. $f$ is monotically increasing on $I$, so that $f(x)\leq f(r_n)$ for all $n$. In other words

$x+x^2\leq r_n-r_n^2$ for all $n$.

Passing to the limit, it is found $x+x^2\leq x-x^2$ which is absurd (if $x\neq 0$). Therefore $I$ does not exist.

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$f'(x) := \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}$

since f(0) = 0,

$f'(0) = \lim_{\Delta x \to 0} \frac{f(\Delta x)}{\Delta x}$

Using $\epsilon - \delta$ method : above equation is equal to (substitute k to $\Delta x$) "for every $\epsilon > 0$, there exists $\delta > 0$ that

$k \in (-\delta,\delta), k \neq 0 \implies \frac{f(k)}k \in (1-\epsilon,1+\epsilon)$."

If k is rational, then $f(k) = k - k^2$, and $\frac{f(k)}k = 1-k$. Therefore, if $\delta = \epsilon$, then $\frac{f(k)}k$ is between $1+\delta$ and $1-\delta$, which is $(1-\epsilon,1+\epsilon)$.

Similar method works if k is irrational ($\delta = \epsilon$)

$\therefore$ whether k is rational or not, for every $\epsilon>0$, there exists $\delta$.

$\therefore$, $f'(0) = 1$.

PS : some error can exists in $\epsilon - \delta$ method

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