Help calculating this limit please

  • #1
Lambda96
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63
Homework Statement
Calculate limit of ##\frac{1}{n+1} |\frac{x-1}{\xi}|^{n+1}## as ##n \rightarrow \infty##
Relevant Equations
none
Hi,

I have problems with task b, more precisely with the calculation of the limit value:

Bildschirmfoto 2024-04-13 um 20.39.51.png


By the way, I got the following for task a ##f^{(n)}(x)=(-1)^{n+1} \frac{(n-1)}{x^n}##

Unfortunately, I have no idea how to calculate the limit value for the remainder element, since ##n## appears in the exponent. I tried it with L'Hôpital's rule and then get ##\Bigl( \frac{x-1}{\xi} \Bigr)^{n+1} \log(\frac{x-1}{\xi})## and if I now form the limit, it would be ##\infty## or not?
 
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  • #2
Why would you apply l’Hopital? It is useful for cases when you have expressions where both numerator and denominator tend to zero.

You should focus on what is being exponentiated: ##(x-1)/\xi##. What are the possible values of this?
 
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  • #3
Thanks Orodruin for your help 👍, I hadn't thought of that, so the amount in the parenthesis is less than 1, which makes the limit at infinity zero.
 
  • #4
Lambda96 said:
so the amount in the parenthesis is less than 1
Can be equal.
 
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  • #5
haruspex said:
Can be equal.
Yes, but the prefactor 1/(n+1) still goes to zero so the limit is still zero. Of course, for the proof to be valid the argumentation should be the correct one though.
 
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  • #6
Orodruin said:
but the prefactor 1/(n+1) still goes to zero so the limit is still zero
Quite so, but I was leaving that to the OP.
 
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