For what values of $x$ does $\lim_{n\rightarrow\infty}\frac{x+x^n}{1+x^n}$ exist?

Question

For what values of $x$ does the following limit exist? $$\lim_{n\rightarrow\infty}\frac{x+x^n}{1+x^n}$$ Here, $n$ is a natural number.

What I have tried so far: The limit exists for $x=1$, since the function simplifies to a constant ($=1$). Further, the limit exists for $x>1$ and the limit equals one in this case. For $x \in (-1,1)$, the limit exists and is equal to $x$.

I am getting confused about the remaining cases ($x\leq-1$). Is the limit undefined for $x\le-1$?

Answer 1

For $|x|<1$ we get $$\lim_{n\to \infty}\frac{x+x^n}{1+x^n}=x$$ For $x=1$ we get $$\lim_{x\to \infty}\frac{x+x^n}{1+x^n}=1$$ For $x<-1$ we get $$\lim_{x\to\infty}\frac{1+x^n}{1+x^n}=1$$ And for $x=-1$ we get $$\lim_{x\to -1}\frac{-1+(-1)^n}{1+(-1)^n}$$ and the limit doesn't exist.

Answer 2

If $x=-1$, $\frac{x+x^n}{1+x^n}$ is $0$ if $n$ is even but undefined if $n$ is odd, so the limit doesn't exist in that case. If $x<-1$, $|x|>1$ so the function is asymptotically $1$, as in the case $x>1$. (This only works if $n$ is restricted to positive integers, so that we obtain the limit of the sequence, unless for $n\notin\Bbb N$ we use a complex definition of $x^n$.)

Answer 3

Rewrite the limit as $$ \lim_{n\to\infty}\frac{x(x^{-n})+1}{x^{-n}+1} $$so that if $|x|>1$, then $|x^{-1}|<1$ whence using the fact that $r^n\to 0$ for $|r|<1$ we have that $$ \lim_{n\to\infty}\frac{x(x^{-n})+1}{x^{-n}+1}=1 $$ for $|x|>1$.

Answer 4

$$\frac{x+x^n}{1+x^n}=1+\frac {x-1}{1+x^n}$$

The only problem point is $x=-1$ for which the denominator is $0$ for odd values of $n$

Otherwise the limit is found with little effort.