Limit and partial derivatives proof

  • Thread starter cezarion
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Homework Statement



Prove that if all partial derivatives up to order [tex]n[/tex] are zero at [tex]\vec{x}[/tex] and [tex]f(x) = 0[/tex] then [tex]\displaystyle\lim_{h \rightarrow 0} \dfrac{f(x + h)}{|h|^n} = 0[/tex]

Homework Equations



[tex]\displaystyle\lim_{h \rightarrow 0} \dfrac{f(x + h) - f(x)}{|h|} = 0[/tex]
[tex]f(x) = 0[/tex]

The Attempt at a Solution



I am not sure if I should be doing this by induction. The first case is trivial. For the general case I am unable to move beyond either the limit [tex]\displaystyle\lim_{h \rightarrow 0} \dfrac{0}{|h|^{k}}[/tex]. I am also not sure how to formulate the equation for the k-th partial derivative. Is it permissible to use the same [tex]h[/tex] for both a given derivative and the next derivative? Thanks for your help.
 

Answers and Replies

  • #2
HallsofIvy
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Are you certain that the first case it "trivial"? Knowing that the partial derivatives are tells you that [tex]\lim_{h\rightarrow 0}\frac{f(\vec{x}+ h\vec{e_i})- f(\vec{x})}{h} = \lim_{h\rightarrow 0}\frac{f(\vec{x}+h\vec{e_i})}{h}= 0[/tex] where h is a real number and [itex]\vec{e_i}[/itex] is the unit vector in the ith coordinate direction. What you want to prove, [tex]\lim_{\vec{h}\rightarrow \vec{0}}\frac{f(\vec{x}+ \vec{h})}{|\vec{h}|^n}= 0[/tex] requires that h be a general vector going to the 0 vector. (It is, of course, very easy to prove one from the other. What is the derivative in direction [itex]\vec{h}[/itex] in terms of the gradient or partial derivatives?)
 
  • #3
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Thank you for the reply. This was actually proved in our textbook, so I am just citing the result in the proof for this problem, leaving just the general case that needs to be proved.
 

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