If $\mid f'(x^{\ast}) \mid <1 $ then $\mid f'(x) \mid I have to show the following: let $ f $ be $C^1$ and let $\mid f'(x^{\ast}) \mid <1 $. Then there exist numbers $ k<1$ and $ \delta>0$ such that $\mid f'(x) \mid <k $ for all $x \in (x^{\ast}-\delta,x^{\ast}+\delta)$. 
Here's my try: by continuity I know that for any $\varepsilon>0$ I can find a delta-neighborhood of $x^{\ast}$, call it $N_{\delta}(x^{\ast})$, such that if $x \in N_{\delta}(x^{\ast})$ then 
$f'(x^{\ast})-\varepsilon< f'(x) <f'(x^{\ast})+\varepsilon$
Then I think I should choose $\varepsilon$ smaller than the distance between $1$ and $f'(x^{\ast})$ and I'm done, but I got a bit lost about how to formalize this last part. Any help appreciated!

We have $|f'(x^{*})|<1$ 
Then there's exists $\varepsilon > 0$ such that $ |f'(x^{*}) | + \varepsilon = 1$
As $f'$ is a continuous function, then we have $lim_{x \to x^{*}}f'(x) = f'(x^{*})$
So, there's exists $\delta > 0$ such that if $x \in (x^{*}-\delta,x^{*}+\delta)$ then $|f'(x) - f'(x^{*})|<\frac{\varepsilon}{2}$ 
Then $-\frac{\varepsilon}{2} < f'(x) - f'(x^{*}) < \frac{\varepsilon}{2}$
Thus $f'(x) < f'(x^{*}) + \frac{\varepsilon}{2} < f'(x^{*}) + \varepsilon = 1$
So if we take $k = f'(x^{*}) + \frac{\varepsilon}{2}$, we are done
