I'm independently working through Boyd and Vandenberghe's "Convex Optimization" and am stuck on the following step in their proof of the first-order convexity condition on page $70$.
If we divide both sides by $t$, we obtain $$f(y) \geq f(x) + \frac{f(x + t(y-x))-f(x)}{t}$$ and taking the limit as $t \rightarrow 0$, yields $$f(y) \geq f(x) + f'(x)(y-x)$$
Specifically, I understand that
$$f'(x) = \lim_{t\rightarrow0} \frac{f(x+t)-f(x)}{t}$$ by definition of a derivative with respect to $t$, but I don't follow how
$$f'(x)(y-x) = \lim_{t\rightarrow0} \frac{f(x + t(y-x))-f(x)}{t}$$
Setting $h = t(y-x)$ and noting that $h\rightarrow0$ as $t\rightarrow0$, we have
$$\begin{align} f'(x) &= \lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h} \\\\ &=\lim_{t\rightarrow0}\frac{f(x+t(y-x))-f(x)}{t(y-x)} \\\\ f'(x)(y-x)&=\lim_{t\rightarrow0}\frac{f(x+t(y-x))-f(x)}{t} \end{align}$$