# Analysis question of limits and continuity [closed]

If $$\{a_n\}_{n=1}^\infty$$ is convergent with limit $$L$$ and $$\{b_n\}_{n=1}^\infty$$ is convergent with limit $$M$$, $$L \gt M$$, and $$c_n = \max(a_n,b_n)$$, then show that $$\lim_{n \to \infty} c_n =L.$$

Exists epsilon >0 s.t. for n>N1, |an-L| < epsilon & Exists epsilon >0 s.t. for n>N2, |bn-M| < epsilon

Set N= max(N1,N2)

I felt s given L>M, beyond a point an>bn, i.e an dominates b, s.t. beyond this value of n. |cn-L| < epsilon

This was my line of thinking. Unsure of how to mathematically demonstrate

• How have you tried to solve this? Where are you stuck? Please edit your question to include this information. Problems that don't demonstrate some effort are usually downvoted and closed. – Robert Shore Apr 13 at 1:07
• If you (@Harsh_Kumar) edit your post to include the info that Robert mentioned, then I would be happy to help. – Matt A Pelto Apr 13 at 2:22
• Since $L > M$, $L-M > 0$. Choose $\epsilon < (L-M)/2$ and see what happens. – marty cohen Apr 13 at 2:42
• You have a good idea with the set "$N=\max(N_1,N_2)$ " bit and your intuition at the end seems appropriate but you seem to be stating the wrong definition for a convergent sequence. The definition says "for every $\varepsilon>0$, there exists a positive integer $N$ such that ...". Notice how the definition says every $\varepsilon>0$ and also that we have $L-M>0$. – Matt A Pelto Apr 13 at 2:46
• Your argument is correct but you should write it out properly. The sequence $a_n-b_n$ tends to a positive number $L-M$ and hence its terms must be positive after a certain point. Thus $\max(a_n, b_n) =a_n$ after a certain value of $n$. – Paramanand Singh Apr 13 at 5:52

Since $$L>M$$, we take $$\varepsilon=\frac{L-M}2$$. Since $$\lim_{n \to \infty} a_n=L$$ and $$\lim_{n \to \infty} b_n=M$$, we know there are positive integers $$N_1$$ and $$N_2$$ such that $$|a_n-L|<\varepsilon \text{ whenever } n \geq N_1 \\\text{and} \\ |b_n-M|<\varepsilon \text{ whenever } n \geq N_2.$$ Select $$N=\max\{N_1, N_2\}$$. So if $$n \geq N$$, then we have $$b_n<\frac{L+M}2 Hence for every $$n\geq N$$ we will have that $$c_n=a_n$$.
Same sort of idea ($$\varepsilon$$) is helpful in some other introductory analysis proofs; for example in showing that the max of two continuous functions is continuous.
• Hint: Either the two functions are equal at $a$ (an arbitrary point in $\mathbb R$ at which you show the max of two continuous functions is continuous) or else there is $\delta>0$ (perhaps the min of 2 deltas) so that one of the two functions remains the max for all $x$ such that $|x-a|<\delta$. There is also a simple formula (somewhat related to this choice of $\varepsilon$) for the max of two functions which could be used. Two birds, one stone :) @HarshKumar – Matt A Pelto Apr 13 at 3:30
• @HarshKumar No. Let $f,g:\mathbb R \to \mathbb R$ be continuous functions, and define $h(x):=\max\{f(x),g(x)\}$. If $f(a)=g(a)$, then for any $\varepsilon>0$ how can we find $\delta>0$ so that $|h(x)-h(a)|=|h(x)-f(a)|=|h(x)-g(a)|<\varepsilon$? If $f(a) \neq g(a)$, then either $f(a)>g(a)$ or $g(a)>f(a)$. First pick an $\varepsilon'>0$ (similar but not identical to the $\varepsilon$ used in the argument above) and find $\delta'>0$ so that the function which is greater at $a$ remains greater for all $x$ s.t. $|x-a|<\delta'$. Then simply pass the continuity of $f$ or $g$ on to $h$. – Matt A Pelto Apr 15 at 20:31
It would be very helpful to know that $$\forall a, b \in \mathbb{R}, \, \max\{a,b\}=\frac{|a+b|+|a-b|}{2}.$$ Then $$c_n = \frac{|a_n+b_n|+|a_n-b_n|}{2} \, \forall n \in \mathbb{N}.$$