Computing the limit of a sequence

Question

From Question:

For each real number x. determine if the sequence $$\left(\frac{1}{{1+x^{n}}}\right)^{\infty}_{n=1}$$ has a limit, and compute it when exist.

For this answer I'm not sure please advice .

Let $a_n = \left(\frac{1}{{1+x^{n}}}\right)$

take limit ; $\lim_{n \to \infty} {a_n} = \lim_{n \to \infty} \frac{1} {{1+x^{n}}} = 0$

Given $\epsilon > 0$

Hence $|a_n - 0|$ = $ |$$\frac{1} {{1+x^{n}}} -0| =\frac{1} {{1+x^{n}}} \leq {\frac{1} {{n}}} \leq {\frac{1} {{N}}} < \epsilon$

Choose $N > \frac{1}{\epsilon}$

$|a_n - L|$ = $ |\frac{1} {{1+x^{n}}}-0| =\frac{1}{{1+x^{n}}} <{\frac{1} {{n}}} <{\frac{1} {{N}}} = {\epsilon}$

=$|\frac{1}{{1+x^{n}}}|<{\epsilon}$, $\forall n \ge N$

it exists when $n \ge\ 1$

Answer 1

For $|x|<1$ we have: $$ \lim_nx^n=0 \Longrightarrow \lim_n\frac{1}{1+x^n}=1. $$ For $|x|>1$ we have $$ \lim_n|x|^n=\infty \Longrightarrow \lim_n\frac{1}{1+x^n}=0. $$ For $x=1$ we have $$ \lim_n\frac{1}{1+x^n}=\frac12. $$ For $x=-1$ the limit does not exist.

Answer 2

Here are some hints.

The sequence involves $x^{n}$.

(1)

Does $\displaystyle \lim_{n\rightarrow\infty} x^{n}$ depend on the value of $x$?

(2)

What happens to $\displaystyle \lim_{n\rightarrow\infty} x^{n}$ when $|x| > 1$? What about $|x| < 1$? $|x| = 1$?

Answer 3

The limit of the sequence depends on the value of $x$. First of all, if $|x|<1$, then $\lim x^n = 0$, thus $\lim a_n = 1$. If $x=1$, then $\lim a_n = 0.5$. If $x>1$, then

$|\frac{1}{1+ x^n}| < \frac{1}{x^n} \to 0$, thus $\lim a_n = 0$. If $x<-1$, then

$|\frac{1}{1+ x^n}| < \frac{1}{|x|^n-1} \to 0$, thus $\lim a_n=0$.