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Show that if $a_n geq b_n$ Then $lim_n a_n ge lim_n b_n$ provided the limit exists.

I'm trying to derive a contradiction and it feels obvious but I can't come up with a clean proof?

Take $c_n=a_n-b_n geq 0$ and let $a=lim a_n$ and $b=lim b_n$. Then we can define $lim c_n$ as follows:

$$lim c_n=lim a_n-lim b_n=a-b$$

Claim: $a-b geq 0$

Suppose $a-b <0$, then $varepsilon =-(a-b)>0$. Since $c_n rightarrow a-b$, so $$(a-b)-varepsilon <c_n <(a-b)+varepsilon$$ for all $n geq K$. In paricular, $$c_K<(a-b)+varepsilon=(a-b)+(-(a-b))=0$$ contradiction to the fact $c_n geq 0$. Thus $a-b geq 0$ and so $lim a_n geq lim b_n$

Here's a hint:

For convenience, define $a=lim_ntoinfty a_n$ and $b=lim_ntoinfty b_n$. Suppose that $a_ngeq b_n$ but $a<b$. Define $epsilon=fracb-a2$. For large enough $n$, we have that

$$|a_n-a|<epsilon$$

Similarly, for large enough $n$, we have

$$|b_n-b|<epsilon$$

Using these inequalities, we can derive a contradiction. What is $a_n$ smaller than? What is $b_n$ larger than?

So looking online, I think that the first problem from

http://www.stat.uchicago.edu/~lekheng/courses/104s09/math104s09-hw4sol.pdf is a pretty good resource.

proof. Let's first define $lim a_n = a$ and $lim b_n = b$. Now for the sake of contradiction, let's suppose that $lim a_n < lim b_n$, or equivalently, $a < b$. We can set $epsilon = fracb-a2>0$.

By the definition of limit, there exists an $N_1$ such that for all $n>N_1$, we have that

$(1)$ $|a_n - a| < epsilon$ $iff$ $a-epsilon < a_n < a + epsilon$

Likewise, there exists an $N_2$ such that for all $n>N_2$, we have that

$(2)$ $|b_n - b| < epsilon$ $iff$ $b-epsilon < b_n < b+ epsilon$.

We can let $N = maxN_1, N_2$. From $(1)$ and $(2)$, we get that for $n> N$

$b_n > b - epsilon = b - fracb-a2 = a + fracb-a2 = a + epsilon > a_n$

which contradicts the hypothesis given. Thus if $a_n ge b_n$, it follows that $lim a_n ge lim b_n$