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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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    @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