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How can I compute the following limit?


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Please avoid double dollar signs in titles. – Pedro Tamaroff Apr 18 '13 at 21:23
up vote 2 down vote accepted

You have to look at different values of $x$. Look at $|x+1|<1$ and $|x+1|\geq 1$ separately. In each case, what can you say about the numerator? Does it go to a finite limit? What can you say about the denominator? Does it blow up? Stay constant? Tend to a finite limit?

You'll have to use that $$|r|<1\implies r^n\to 0$$ $$|r|> 1\implies |r^n|\to \infty$$ $$r=1\implies r^n\to 1$$ $$\forall r>0\; ; \; r^{1/n}\to 1$$

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$$\lim_{n\to\infty}\left({\sqrt[n]{|x-1|}+1\over (x+1)^n+1}\right)\to{\lim_{n\to\infty}\left({\sqrt[n]{|x-1|}+1}\right)\over\lim_{n\to\infty}\left({(x+1)^n+1}\right)}$$ $\lim_{n\to\infty}\left({(x+1)^n+1}\right)$ is only defined if x>-1, where it is infinity, so the limit is 0 becuase any number over infinity is 0;

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No, what you're stating is false. When $|x+1|<1$ the denominator goes to $1$, since $(1+x)^n$ goes to $0$. It is true however, we should take care when $(x+1)^n+1=0$. This can only happen when $n$ is odd, and in such a case what we want to avoid is $x=-2$. – Pedro Tamaroff Apr 18 '13 at 22:40
Oops, I forgot about that first part. I suppose that means if -1<x<0, then the limit is infinity and if x=0 then the limit is 1. Also, what you say is true and that's why the limit is undefined if x<-1: because infinity is neither odd nor even, so the limit diverges. – hacatu Apr 18 '13 at 23:01
If $|x+1|>1$ the limit is $0$. The denominator might take arbitrarily large values, both positive and negative, but there is no problem with that. – Pedro Tamaroff Apr 18 '13 at 23:02
how is the limit 1 if x=0? You would get [(-1)^(1/n)+1]/2, and for odd values of n you would get 0. – Ovi Apr 19 '13 at 3:39

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