The question:

Let ($a_n$) be a sequence such that $$\lim _ {n\to\infty} a_n=a$$ Let ($b_n$) be a sequence such that $$(b_n)=\max(a_n,(a_{n})^2)$$
For all $a>0$, calculate $\lim\limits_{n\to\infty}b_n$

My solution seems too easy and therefore I think I'm missing something here.

From limit arithmetic, $\lim\limits_{n\to\infty}(a_n)^2=\lim\limits_{n\to\infty}(a_n)*\lim\limits_{n\to\infty}(a_n)=a*a=a^2$. From the definition of $b_n$ I can say that $\forall\space a>1$, $b_n=(a_n)^2$. Therefore, for $n$ large enough, $b_n=(a_n)^2$.

$$\Rightarrow \lim_{n\to\infty}b_n=\lim_{n\to\infty}(a_n)^2=a^2$$

Am I right?


  • $\begingroup$ Your proof is right under hypothesis $a>1$. What if $a\leq 1$? $\endgroup$ – Hamou Sep 12 '15 at 9:57
  • $\begingroup$ @Hamou I certainly understand, so what do I do in that case? split to cases? $\endgroup$ – Alan Sep 12 '15 at 9:58
  • $\begingroup$ The case $a<1$: $a_n^2\leq a_n$ for large $n$...and $b_n=a_n$. The case $a=1$,. $\endgroup$ – Hamou Sep 12 '15 at 10:02

$$f(x)=\max(x,x^2)$$ is a continuous function on $\mathbb{R}$, hence: $$ \lim_{n\to +\infty} b_n = \lim_{n\to +\infty} f(a_n) = f\left(\lim_{n\to +\infty} a_n \right) = \color{red}{\max(a,a^2)}.$$

  • $\begingroup$ So the answer is $a^2$ but the way I did it is wrong? $\endgroup$ – Alan Sep 12 '15 at 9:57
  • 1
    $\begingroup$ It is $a^2$ only when $a\geq1$. So no, your answer was wrong. $\endgroup$ – Hasan Saad Sep 12 '15 at 9:59
  • $\begingroup$ @Alan: the answer is not $a^2$, the answer is $\max(a,a^2)$. That equals $a$ if $a\in(0,1)$, not $a^2$. You assumed $a>1$ but that was not given by the statement. $\endgroup$ – Jack D'Aurizio Sep 12 '15 at 10:00
  • $\begingroup$ Thank you, I didn't know that there's a possibility for that kind of answer to a question, that's where I made the mistake, now I will know for the next time. Thank you! $\endgroup$ – Alan Sep 12 '15 at 10:06
  • $\begingroup$ @Alan: you're welcome. $\endgroup$ – Jack D'Aurizio Sep 12 '15 at 10:06

$\max(\alpha,\beta)=\dfrac{|\alpha-\beta|+\alpha+\beta}{2}$. hence $b_n=\max(a_n,a_n^2)=\dfrac{|a_n-a_n^2|+a_n+a_n^2}{2}\to \dfrac{|a-a^2|+a+a^2}{2}=\max(a,a^2)$.


You are right but need to make it a bit formal for aproof.

For $a>1$ you can say $\exists N$ such that $\forall n>N$ $|a_n-a|<\epsilon \implies a_n>a-e$

choose $\epsilon$ so that $a-\epsilon>1$.

so $\forall n>N b_n=a_n^2$ the rest is like you did so $lim_{n\rightarrow \infty}b_n=a^2$. similarly you can do for $a<1$ then the answer will be $a$.

  • $\begingroup$ ... and get that for $a<1$ the answer is not $a^2$, so the OP's answer is not right. $\endgroup$ – Jack D'Aurizio Sep 12 '15 at 10:01

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