# Could someone explain this proof to me?

I found this form MIT OCW

http://ocw.mit.edu/courses/mathematics/18-101-analysis-ii-fall-2005/lecture-notes/lecture3.pdf

Please scroll down to Theorem 2.4: If f is differentiable at a, then for every u the directional derivative of f in the direction of u at a exists

Could someone explain (2.17)? How did that go to $0$? And I don't see how 2.18 follows from that? Did they multiply and divide t in 2.17?

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$(2.17)$ is another way of writing the expression in $(2.16)$, which goes to $0$ as $t \rightarrow 0$.

Then you have the product of two things going to $0$ as $t \rightarrow 0$, where one of the things is $\frac{t}{|t|} \frac{1}{|u|}$ and the other is the stuff in parentheses.

Note that the first thing does not approach $0$ as $t \rightarrow 0$, so it must be the parenthetical stuff going to $0$.

Since the parenthetical thing is of the form $X - Y$, this means that $X - Y \rightarrow 0$ as $t \rightarrow 0$, i.e., $X = Y$ as $t \rightarrow 0$.

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How do they know the thing $X - Y$ goes to $0$? That's not obvious at all, especially with the $\mathbb{u}$ in there – Hawk Jan 24 '13 at 5:05
if $\lim_{t \rightarrow 0} \alpha(t) \cdot \beta(t) = 0$, then either $\alpha = 0$ or $\beta = 0$ or both as $t \rightarrow 0$. In this case, you have the product of two things, where one of them is not going to $0$ as $t \rightarrow 0$. Thus, it must be the other one that goes to $0$. – Benjamin Dickman Jan 24 '13 at 5:12
Sorry, I am asking why does (individually, as a limit) $X - Y$ go to $0$?. I want to know why $\alpha \to 0$ – Hawk Jan 24 '13 at 5:13
$(2.16)$ gives you an expression of the form $\lim_{t \rightarrow 0} \gamma(t) = 0$. Next, the author rewrites $\gamma(t) = \alpha(t) \cdot \beta(t)$, so that this latter expression will also go to $0$ as $t$ does. Since $\alpha(t)$ does not go to $0$, it must be that $\beta(t) = X - Y \rightarrow 0$ as $t \rightarrow 0$, i.e., $X = Y$ as $t \rightarrow 0$. – Benjamin Dickman Jan 24 '13 at 5:17
Okay sorry, I meant to say why does $\beta (t)$ go to $0$. There is a miscommunication. Let's just forget the last step is even there, that is "$\to 0$". I want to know where does $X - Y \to 0$. I know the process of elimination here – Hawk Jan 24 '13 at 5:23