Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 2 down vote favorite

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?

share|improve this question

1 Answer

up vote 1 down vote accepted

$(2.17)$ 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 $t \rightarrow 0$.

share|improve this answer
How do they know the thing $X - Y$ goes to $0$? That's not obvious at all, especially with the $\mathbb{u}$ in there – sidht Jan 24 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 $0$. – B.D Jan 24 at 5:12
Sorry, I am asking why does (individually, as a limit) $X - Y$ go to $0$?. I want to know why $\alpha \to 0$ – sidht Jan 24 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 $t \rightarrow 0$. – B.D Jan 24 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 – sidht Jan 24 at 5:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.