How to go from limit of vector norm to 'normal' limit

Berrius
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This is not really a homework question, but I've come across this while preparing for a test

Homework Statement


Let f:U \subseteq R^n -> R^m be a function which is differentiable at a \in U, and u \in R^n
It is then stated that it is clear that:
lim_{t \to 0} \frac{||f(a+t*u)-f(a)-D_f(t*u)||}{||t*u||} = 0 => lim_{t \to 0} \frac{f(a+t*u)-f(a)-D_f(t*u)}{|t|} = 0

How do they get this result?

The Attempt at a Solution


I've tried using the epsilon/delta-def of limits but where the first limit is about real numbers, the second is about vectors in the R^m, so I am totally confused.
 
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Denominator: ##||t \cdot u|| = |t| \cdot ||u||## and the limit is linear in ##||u||##.
Numerator: The norm of a vector can go to 0 if and only if the vector itself goes to 0.
 
I know ofcourse ||v||=0 iff v=0, but why is this still true when i put a limit in front of it?
 
You can show it for each component in some basis, for example. They all go to 0.
 
It is the definition that vector v converges to vector w if the norm of their difference converges to 0.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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