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I failed to understand how what do author of my vector calculus textbook arrived at below equation:

Based on equation below: $$\lim_{x\to a}\frac{\|f(x)-[f(a)+Df(a)(x-a)]\|}{\|x-a\|}=0$$

Given $\|f(x)-f(a)-Df(a)(x-a)\|$
Thus since f is differentiable at a, we can make $\|f(x)-f(a)-Df(a)(x-a)\|$
as small as we can wish by keeping $\|x-a\|$ appropriately small.
In particular,$$\|f(x)-f(a)-Df(a)(x-a)\|\leq\|x-a\|$$ if $\|x-a\|$ is appropriately small.

Note: I modify above proof from my textbook to fit the context of this question.
Note that the 'D' above is differential symbol.
to be more precise with my question, I was thinking straightforward
In first equation 0 multiply with $\|x-a\|$ should be 0
I'm not sure how the author arrive with $\leq\|x-a\|$ at the last equation.

1 Answers 1

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$$\lim_{x\to a}\frac{\|f(x)-[f(a)+Df(a)(x-a)]\|}{\|x-a\|}=0$$ So you can find $\varepsilon>0$ such as $\|x-a\|<\varepsilon\Rightarrow\frac{\|f(x)-[f(a)+Df(a)(x-a)]\|}{\|x-a\|}\le 1$.

So you have well $\|f(x)-f(a)-Df(a)(x-a)\|\leq\|x-a\|$ if $\|x-a\|$ is appropriately small.

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    thanks.But is it correct if i interpret this as $\|x-a\|<\delta=\epsilon$ ? because normally definition of limit is defined as $0<\|x-a\|<\delta$ and $\|f(x)-L\|\leq\epsilon$2012-06-16
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    by definition you are given an $ \epsilon $ and you are finding the $ \delta $ in order to have $|\!| f(x) - L |\!|$ so there are times that the $\delta $ that does your works is equal to $\epsilon $.2012-06-16
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    With your notation : use $\varepsilon=1$ and your $\delta$ will be my $\varepsilon$...2012-06-16
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    @clark,@JBC I remember looking at some graph on definition on with limit with delta and epsilon.It made sense now when putting everything together.Thanks guy for your help.2012-06-16