Suppose $\nabla f (x) \cdot d < 0$, prove that there exists $\delta > 0$ such that $$f(x + \tau d ) < f(x)$$ for all $\tau \in (0, \delta)$
My proof consists of only a few lines.
Since $$L =\lim_{\tau \to 0} \frac{f(x+ \tau d) - f(x)}{\tau} < 0,$$ then there exists $\delta > 0$ (since the limit exists) for $\tau \in (0, \delta) \implies f(x + \tau d) - f(x) <L+\epsilon.$
Here we will take $\epsilon = -L$, giving us the desired inequality.
Now in my notes, there is another slightly more messy way, which there is one step I cannot follow.
Since $$\lim_{\tau \to 0} \frac{f(x+ \tau d) - f(x)}{\tau} < 0,$$ then there exists $\epsilon > 0$ such that $$\lim_{\tau \to 0} \frac{f(x+ \tau d) - f(x)}{\tau} - \epsilon < 0$$ and there exists $\bar{\tau} > 0$ such that $\tau \in (0, \bar{\tau} ) \implies $
$$-\frac{3\epsilon}{2} = - \epsilon - \frac{\epsilon}{2} < \frac{f(x+ \tau d) - f(x)}{\tau} < - \epsilon + \frac{\epsilon}{2} = - \frac{\epsilon}{2}< 0$$
So there exists $\bar{\tau} > 0$ such that $\tau \in (0, \bar{\tau} ) \implies f(x+ \tau d) > < f(x) - \frac{\tau \epsilon}{2} < f(x)$
I am guessing the middle step is actually trying to use the second limit, but I am guessing they really mean to say $$\lim_{\tau \to 0} \frac{f(x+ \tau d) - f(x)}{\tau} + \epsilon < 0$$
Am I correct or did I really just copied the entire proof wrong?