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Does there exists a differentiable function $f: \mathbb{R} \to \mathbb{R}$ with $f'(0)=0$ and the existence of a sequence $(x_n)_n$ in $\mathbb{R}$ such that $x_n \to 0$ implies $f(x_n)\to \infty$.

Attempt I guess not, because the derivative is bounded at $x=0$ we can have an open interval $(-\epsilon,\epsilon)$ around $x=0$ on which the function is uniformly continuous, and thus also bounded?

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Because $f$ is differentiable at $0$ (note that we do not even need that $f$ is differentiable or even continuous elsewhere), we have that $\lim_x\to 0{f(x)-f(0)}{x}$ exists, hence with $\epsilon:=1$ there exists $\delta>0$ such that $$\left|\frac{f(x)-f(0)}{x}-f'(0)\right|<1 $$ for $|x|<\delta$. In paticular, $f(x)$ is between two lines with slopes $f'(0)\pm1$ through $(0,f(0))$.

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  • $\begingroup$ Thank you for the answer, your argumentation allows me to have a more visual idea of the situation! $\endgroup$ – Sim Aug 22 '19 at 13:21
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In order for $f$ to be differentiable in $0$, it has to be continuous in $0$. Therefore $$\lim_{n\rightarrow\infty}f(x_{n})=f(0)$$ if $\lim_{n\rightarrow\infty}x_{n}=0$.

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  • $\begingroup$ Nice simple solution. $\endgroup$ – zhw. Aug 22 '19 at 19:16

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