Does L'Hopital's Rule Apply to Limits of x^(1-p) as x Approaches Infinity?

[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

 

[dumbQuestion]

I'm also curious about the answer


lim(x-->∞) x^1-p = lim(x-->∞) x¹x^-p = lim(x-->∞) x/x^p = ∞/∞

So we apply L'hopitals:

lim(x-->∞) 1/(px^p-1) = 1/p lim(x-->∞) 1/(x^p-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

 

[Vorde]

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.[/color]

 

[dextercioby]

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&gt;0

Which should be easier to handle when considering the limit.