Strichartz, in his book The Way of Analysis, says that,
Suppose $f$ is a real valued function defined in an open neighborhood of $x_0$, where $x_0$'s neighborhood is subset of $\Bbb R$, and $f$ is differentiable at $x_0$. Define $g(x)=f(x_0)+f'(x_0)(x-x_0)$ and $g_1(x)=f(x_0)+a_1(x-x_0)$, where $a_1\ne f'(x_0)$. Then, for any $m\in \Bbb N$ $$\begin{align}|f(x)-g_1(x)| &=|(f(x)-g(x))+(g(x)-g_1(x))|\\ &\ge |g(x)-g_1(x)|-|f(x)-g(x)|\\ & \ge |a_1-f'(x_0)||x-x_0|-\frac 1m|x-x_0|\end{align}$$ if $|x-x_0|\le \frac 1n$ for suitable $n\in \Bbb N$, and so if we take $\frac 1m< \frac{|a_1-f'(x_0)|}{2}$, we have $$|f(x)-g_1(x)|\ge |a_1-f'(x_0)||x-x_0|/2$$ for all $x$ satisfying $|x-x_0|\le \frac 1n$ for suitable $n$.
But, later he says that for any affine function $g_1(x)=f(x_0)+a_1(x-x_0)$, where $a_1\ne f'(x_0)$, we have $f(x)-g_1(x)=O(|x-x_0|)$ as $x\rightarrow x_0$, that is there exists $\frac1n$ and $c>0$ such that $|x-x_0|<\frac 1n$ implies $|f(x)-g_1(x)|\le c|x-x_0|$.
How can both be possible?
Why is this bothering you? You have $$|a_1-f'(x_0)||x-x_0|/2\le |f(x)-g_1(x)|\le c |x-x_0|$$ and so $c\ge |a_1-f'(x_0)|/2$. If you call $h(x)=f(x)-g_1(x)$ you have that $h(x_0)=0$ and $h'(x_0)\ne 0$, so you are saying that near $x_0$ the function $h$ behaves like the line $y=h'(x_0)(x-x_0)$.