2
$\begingroup$

I am trying to find the limit of the sequence,

$$f_n(x)=\frac{x^n}{n}$$

for $x\in [0,1]$

I am just wondering if it is $f(x)=0$ because when $x=1$ then $\frac{x^n}{n}=\frac{1}{n}=0$ and when it is $x\leq 1$ the numerator goes to $0$ as $n$ goes to infinity.

Or is there a more mathematical approach to this result?

$\endgroup$
  • $\begingroup$ Prove that $\forall x \in [0,1]: \displaystyle \lim_{n \to \infty} f_n(x) = 0$ $\endgroup$ – DHMO Apr 16 '17 at 14:10
  • $\begingroup$ @DHMO how do I go about that? $\endgroup$ – fr14 Apr 16 '17 at 14:11
  • $\begingroup$ How deep do you need? Do you need to do it epsilon-delta style? $\endgroup$ – DHMO Apr 16 '17 at 14:13
  • $\begingroup$ I was showing that the sequence converges uniformly but first I wanted to find its limit, showing that the sequence converges uniformly is easy but shouldn't I find its limit first? $\endgroup$ – fr14 Apr 16 '17 at 14:14
  • $\begingroup$ The limit is $0$. One way to see it: for each $x$, this sequence is a multiplication of a bounded sequence ($x^n$) and a sequence whose limit is $0$ ($1/n$) $\endgroup$ – 35T41 Apr 16 '17 at 14:15
1
$\begingroup$

Yes, in fact the function $x \mapsto 0$ on $[0,1]$ is the uniform limit of the sequence $(f_{n})$. To see this, note that $x \in [0,1]$ and $n \in \mathbb{N}$ imply $$ \frac{x^{n}}{n} \leq \frac{1}{n}. $$ Note that $1/n \to 0$; so $f_{n} \to 0$ on $[0,1]$ uniformly.

If you want an epsilon-analysis: If $\varepsilon > 0$, then there is some $N \in \mathbb{N}$ (say $N := \lfloor \frac{1}{\varepsilon} \rfloor + 1$) ensuring $n \geq N$ implies $1/n < \varepsilon$. So $n \geq N$ implies $|\frac{x^{n}}{n}-0| = \frac{x^{n}}{n} \leq \frac{1}{n} < \varepsilon$ for all $x \in [0,1]$; this shows that $(f_{n})$ converges uniformly on $[0,1]$ to the function $x \mapsto 0$.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Let $t\in [0,1]$.

Note that $0 \leq t^n \leq 1$, and therefore $0 \leq t^n/n \leq 1/n$. Take the limit and use the sandwich's theorem.

| cite | improve this answer | |
$\endgroup$

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.