0
$\begingroup$

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 $n \ge\ 1$

$\endgroup$
1
  • $\begingroup$ Your proof is a little messy, so I have some questions - are you trying to show that the limit of the sequence is zero for any real number $x$? $\endgroup$ Oct 14 '13 at 7:43
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $|x| < 1$? $|x| = 1$?

$\endgroup$
0
$\begingroup$

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 $\lim a_n=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.