Can this limit be computed?

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  • #1
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I've been trying for a while to compute this limit. Is there even a unique solution to this problem?

[tex] \lim_{x→∞} x^{1-p} [/tex] where [itex] p>1 [/itex]

I tried using L'Hopital's rule, but it didn't work out.

BiP
 

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  • #2
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I'm also curious about the answer


lim(x-->∞) x1-p = lim(x-->∞) x1x-p = lim(x-->∞) x/xp = ∞/∞

So we apply L'hopitals:

lim(x-->∞) 1/(pxp-1) = 1/p lim(x-->∞) 1/(xp-1)


so as long as p-1>0 the limit goes to 0, right? And we know, p>1, so we know that p-1>0 so this should go to 0
 
  • #3
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It seems to me that, intuitively, your function should approach zero regardless of p (assuming p > 1). Let me see what I can do more legitimately though.

First split ##x^{1-p}## into ##x^1 x^{-p}##


From there I would make it a quotient and try some fancy l'hopital's on it. I'd help more but I need to get somewhere. Good luck, however!

Mod note: in LaTeX expressions with exponents with more than one character, use braces - {} - around the exponent. I fixed the exponents above.
 
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  • #4
dextercioby
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I've been trying for a while to compute this limit. Is there even a unique solution to this problem?

[tex] \lim_{x→∞} x^{1-p} [/tex] where [itex] p>1 [/itex]

I tried using L'Hopital's rule, but it didn't work out.

BiP
1. Do you know in which cases you're allowed to use L'Hôpital's rule ??
2. By [itex] \infty [/itex] do you assume [itex] +\infty [/itex] ?
2. If p>1, then 1-p <0 = -s, s>0, so that the object under the limit becomes

[tex] \frac{1}{x^s}, ~ s>0 [/tex]

Which should be easier to handle when considering the limit.
 
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