## Can this limit be computed?

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

$$\lim_{x→∞} x^{1-p}$$ where $p>1$

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

BiP
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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
 Recognitions: Gold Member 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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## Can this limit be computed?

 Quote by Bipolarity I've been trying for a while to compute this limit. Is there even a unique solution to this problem? $$\lim_{x→∞} x^{1-p}$$ where $p>1$ 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 $\infty$ do you assume $+\infty$ ?
2. If p>1, then 1-p <0 = -s, s>0, so that the object under the limit becomes

$$\frac{1}{x^s}, ~ s>0$$

Which should be easier to handle when considering the limit.

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