# Does such a differentiable function exist?

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?

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 $$\lim_{n\rightarrow\infty}x_{n}=0$$.
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 $$(0,f(0))$$.