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Suppose $(a_n)$ is a sequence and $\lim_{n\to\infty} a_n = a$ and let $(b_n)=(\max\{a_n,a_n^2\})$. I have to prove/disprove that:

  • If $a>1$ then $\lim_{n\to\infty} b_n = a^2$

  • If $a=1$ then $\lim_{n\to\infty} b_n = 1$

I believe both are true, but I fail to prove it, although it looks intuitive and easy. Any hints?

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$$ \lim_nb_n=\max\{a,a^2\}=\begin{cases}a^2 & \mbox{ if } a>1\\ 1 &\mbox{ if } a=1\end{cases}. $$


Added to the proof Since $$ \max\{x,y\}=\frac{x+y+|x-y|}{2} \quad \forall x,y\in \mathbb{R} $$ we have \begin{eqnarray} |\max\{x_1,y_1\}-\max\{x_2,y_2\}|&=&\frac12\left|x_1+y_1+|x_1-y_1|-x_2-y_2-|x_2-y_2|\right|\\ &\le&\frac12|x_1-x_2|+\frac12|y_1-y_2|+\frac{\left||x_1-y_1|-|x_2-y_2|\right|}{2}\\ &\le& \frac12|x_1-x_2|+\frac12|y_1-y_2|+\frac{|x_1-y_1-x_2+y_2|}{2}\\ &=& \frac12|x_1-x_2|+\frac12|y_1-y_2|+\frac{|x_1-x_2-(y_1-y_2)|}{2}\\ &\le& \frac12|x_1-x_2|+\frac12|y_1-y_2|+\frac12|x_1-x_2|+\frac12|y_1-y_2|\\ &=&|x_1-x_2|+|y_1-y_2|. \end{eqnarray} Therefore, if $\lim_nx_n=x$ and $\lim_ny_n=y$, then $$ \lim_n|\max\{x_n,y_n\}-\max\{x,y\}|\le \lim_n\left[|x_n-x|+|y_n-y|\right]=0, $$ i.e. $\lim_n\max\{x_n,y_n\}=\max\{x,y\}$.

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  • $\begingroup$ I think he's looking for a justification that you can bring the limit inside the max, i.e. $\lim_{n\to\infty} \max(a_n, a_n^2) = \max(\lim_{n\to\infty} a_n, \lim_{n\to\infty} a_n^2)$ $\endgroup$ – Andre Apr 18 '15 at 17:45
  • $\begingroup$ @Andre - indeed. $\endgroup$ – user232495 Apr 18 '15 at 17:55
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For $a > 1$:

By the limit definition of $\lim_{n\to\infty}a_n=a$, there is an $N \in \mathbb N$ with $$n > N \Rightarrow |a_n-a| < \epsilon.$$ for all $\epsilon > 0$. Let $\epsilon = \frac{a-1}{2}$, then $1 < a-\epsilon$, and therefore, for all $n > N: a_n > 1$. From this we know, again for all $n > N$, that $a_n^2 > a_n$. Thus, we can conclude that $\lim_{n\to\infty} \max\{a_n, a_n^2\} = \lim_{n\to\infty} a_n^2 = a^2$.

The other case can be solved similarly.

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Suppose $(a_n)$ is a sequence and $\lim_{n\to\infty} a_n = a$.

Then you know by algebra of limits that $\lim_{n\to\infty} a_n^2 = a^2$. - Fact 1. This is a basic fact, so you can look it up on the internet, but I think you should have been taught that.

Now suppose $a>1$. So $a-1 = \varepsilon, \exists \varepsilon > 0$. Then there exists an $M_1$ such that for all $n>M_1$, $a - |a_n| = |a| - |a_n| \leq||a| - |a_n|| \leq |a - a_n| < \varepsilon = a-1$.

So $- |a_n| < - 1$ hence $|a_n|>1$ for all $n>M_1$.

Now $\max\{a_n,a_n^2\} = a_n^2$ since $|a_n|> 1$ for all $n>M_1$. Now we know from fact 1 that for all $\delta > 0$, there exists an $M_2$ such that for all $n>M_2$, $|a_n^2 - a^2|<\delta$. Let $M = \max\{M_1,M_2\}$. Sofor all $n>M$, $|\max\{a_n,a_n^2\} - a^2| = |a_n^2 - a^2| <\delta$

Now suppose $a = 1$, both sequences, $a_n^2$ and $a_n$ tend to $1$ by Fact 1. Now that means for all $\epsilon >0$, there exists $N_1$ such that for all $n>N_1$, $|a_n - 1| < \epsilon$ and there exists $N_2$ such that for all $n>N_2$, $|a_n^2 - 1| < \epsilon$. Take $N:=\max\{N_1,N_2\}$. So there exists $N$ such that for all $n>N$, $|a_n - 1| < \epsilon$ and $|a_n^2 - 1| < \epsilon$, hence $|\max\{a_n,a_n^2\} - 1| < \epsilon$ .

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