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For the function $$f(x)=\lim_{n\to \infty}\;\large \frac {1} { \frac 1 {x^n} +1},$$ how do I show that the point $f(-1)$ does not exist?

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How about induction? Show that for two adjacent points in the sequence 1,2,3,.., n, n+1 that adjacent points are a finite distance apart. – Tpofofn Jan 18 '13 at 1:00
You might want to make that more precise, @Tpofofn. Subsequent sequence elements being "a finite distance apart" needn't guarantee a failure to converge. Consider the constant functions $f_n(x)=\frac1n$ for $n=1,2,3,....$ It can be said for any $x$ and any $n$ that $f_n(x)$ and $f_{n+1}(x)$ are "a finite distance apart", but clearly $\lim_{n\to\infty}f_n$ is the zero function. – Cameron Buie Jan 18 '13 at 1:04
By plugging in $x = -1$ and noting the denominator of the expression is $0$. – Benjamin Dickman Jan 18 '13 at 1:27

A function being defined or not defined is not the same thing as a limit not existing.

That is, $f(x) \neq \dfrac 1 {\dfrac 1 {(x)^n} +1} = 1 - \dfrac{1}{(x)^n + 1}$. Clearly the function to the right is not defined at $-1$.

Some functions are undefined at certain values but nonetheless, limits exist.

In this case, the function in question is defined as a limit: That is, the function in question is not the expression $\dfrac 1 {\dfrac 1 {(x)^n} +1} = 1 - \dfrac{1}{(x)^n + 1}$, but rather the limit as $n\to \infty$ of that expression at the value $x$: $$f(x) =\quad\lim\limits_{n\to \infty}\dfrac 1 {\dfrac 1 {(x)^n} +1} = \lim_{n\to \infty} 1 - \dfrac{1}{(x)^n + 1}$$ which is a limit that exists for some values $x$, but not for $-1$: That is, $f(-1)$, the limit when $x = -1$, does not exist.

(1) $\lim\limits_{n\to \infty}\dfrac 1 {\dfrac 1 {(-1)^n} +1} \;=\; \lim_{n\to \infty} 1 - \dfrac{1}{(-1)^n + 1}\quad$ does not exist.

You need only explain why the limit does not exist: as $n \to \infty$, for odd $n$, the limit diverges to infinity, for even $n$, it approaches $1$, oscillating, hence failing to exist.

Since $f(x)$ by is defined $\quad\lim\limits_{n\to \infty}\dfrac 1 {\dfrac 1 {(x)^n} +1} = \lim_{n\to \infty}\quad 1 - \dfrac{1}{(x)^n + 1}$,

it must therefore follow from (1) that $f(-1)=\lim\limits_{n\to \infty}\dfrac 1 {\dfrac 1 {-1^n} +1},\;$ fails to exist.

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Yes, it fails to exist. +1 – S. Snape Jan 31 '13 at 14:45
Duncan - was this all clear enough for you! – amWhy Jan 31 '13 at 14:47
I keep + for another day. It seems that it fails for me too. ;( – S. Snape Jan 31 '13 at 14:47
@Babak - yeah, I've "maxed out" or "capped" the 200 up-vote rep for the day. (Until ~ 9 hours from now!) – amWhy Jan 31 '13 at 14:49

Note that $$\cfrac1{\cfrac1{x^n}+1}$$ fails to be defined at $x=-1$ for all odd $n$. There are infinitely many such $n$, so there cannot be a limit. (Use $\delta$-$\epsilon$ definition of sequence convergence. Pay special attention to the quantifiers.)

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For n odd $(-1)^n=-1$ and $\dfrac 1 {\dfrac 1 {x^n} +1}$ is not well defined or you can think $\dfrac 1 {\dfrac 1 {(-1)^n} +1}=\infty$ for n odd. So $\dfrac 1 {\dfrac 1 {x^n} +1}$ is oscillating between $\infty$ and $\frac{1}{2} $

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