Analysis Problem on Differentiability

Prove that there is no differentiable function $f(x)$ defined on $(-\infty, \infty)$ such that $f'(0) = 1$, but $f'(x) \geq 2$ for $x \ne 0.$

So I use contradiction method, suppose there exists a function $f(x)$ with those properties, using the definition of limit gives me

$f'(0) = \lim_{h\rightarrow0} \mid \frac {f(0+h)-f(0)}{h}\mid$=1 and $f'(x_0) = \lim_{ h\to 0} \mid \frac {f(x_0+h)-f(x_0)}{h}\mid$ $\geq 2$ for $x_0 \in R$ but $x_0\neq 0.$ I don't know how to come up with a contradiction. Please help.