GRE math subject test prep

In summary, the conversation discusses the conditions for a differentiable function with a finite limit of its first and second derivatives as x approaches infinity. It is mentioned that if the limit of the first derivative is zero, then the limit of the second derivative must also be zero, but an example is requested to demonstrate when this may not be the case. The example given is g(x)=f'(x), with the function sin(x^2)/x^2 as a possible counterexample.
  • #1
terhorst
11
0
#34 on the much-discussed http://ftp.ets.org/pub/gre/Math.pdf" :

Suppose [tex]f[/tex] is a differentiable function with [tex]\lim\limits_{x \to \infty }f(x)=K[/tex] and [tex]\lim\limits_{x \to \infty }f'(x)=L[/tex] for some [tex]K,L[/tex] finite. Which must be true?
  1. [tex]L=0[/tex]
  2. [tex]\lim\limits_{x \to \infty }f''(x)=0[/tex]
  3. [tex]K=L[/tex]
  4. [tex]f[/tex] is constant.
  5. [tex]f'[/tex] is constant.

Answer is 1. Is this because [tex]f[/tex] might be [tex]C^1[/tex]? Can you give an example of a function where the limit of the first derivative exists but the limit of the second derivative is not zero? Thanks!
 
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  • #2
I am a little befuddled by this. If 1. is true, it seems like 2. must also be true.
Let [tex]g(x)=f'(x)[/tex]
We know
[tex]\lim_{x\rightarrow\infty} g(x) = K = 0 [/tex]
So it should follow that
[tex]\lim_{x\rightarrow\infty} g'(x) = \lim_{x\rightarrow\infty} f''(x) = L = 0 [/tex]
 
  • #3
If the limit of the second derivative exists then it is zero. But it may not exist - even if the function is C^2. Try sin(x^2)/x^2.
 

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