Prove that if one function is greater or equal than another, then so it is its limit.

Question

Let $g(x)$ and $f(x)$ be real functions such that $f(x) \le g(x); \forall x$ such that both $f$ and $g$ are defined. Using the $\epsilon$-$\delta$ definition of limit show that $\lim_{x \to a} f(x) \le \lim_{x \to a} g(x)$, given that both are defined.

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Here's my solution:

Let's assume the contrary, so let $\lim_{x \to a} f(x) = M$ and $\lim_{x \to a} g(x) = N$, such that $M>N$. Now using the condition we have:

$$f(x) + N < g(x) + M: \quad \quad \forall x \in D_f \text{ and } x \in D_g$$ $$f(x) - M < g(x) - N; \quad \quad \forall x \in D_f \text{ and } x \in D_g$$

Hence there always exists $\epsilon$ such that $\mid f(x) - M \mid < \epsilon <\mid g(x) - N \mid$ or $\mid f(x) - M \mid > \epsilon > \mid g(x) - N \mid$, since the set of reals is dense and we don't have equality.

Hence for this $\epsilon$ we're not able to find $\delta$ such that:

$$0<\mid x-a \mid < \delta \implies \mid f(x) - M \mid < \epsilon \quad \text{or} \quad 0<\mid x-a \mid < \delta \implies \mid g(x) - N \mid < \epsilon$$

which means that one of the limits is wrong which isn't possible. Hence $M\le N$. Q.E.D.

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I think this proof is alright, but is it possible to prove the statement without the initial assumption, or at least without using the fact that the set of reals is dense?

Answer 1

I didn't check your proof, but here is a much easier way to prove your result.

Let $$L_1:=\lim_{x\to a}f(x)\quad\text{and}\quad L_2=\lim_{x\to a}g(x).$$ Let $(x_n)$ be a sequence that converges to $a$. For all $n$ we have, $$ f(x_n)\leq g(x_n),$$ and thus $$L_1=\lim_{n\to\infty }f(x_n)\leq \lim_{n\to\infty }g(x_n)=L_2,$$ which proves the claim.

Answer 2

Definitely have to apply the density property, how else will you be able to justify that if $a<b$ then there exists $c$ such that $a<c<b$

Counter example: Let $f(x)\equiv M, g(x)\equiv N$ be constant functions. Then $|f(x)-M|<|g(x)-N|$ is violated.

Answer 3

Repeating what I said in the comments: I am inclined to say this proof is false. If $x\to a$ then both $\vert f(x)-M\vert\to0 $ and $\vert g(x)-N\vert\to0$. This would mean that $ \vert f(x) - M \vert < \epsilon <\vert g(x) - N \vert$ does not hold when $x\to a$. Your proof does not hold, because the epsilon you choose is nonexistent for certain values of $x$ (in particular for the values we're interested in). We cannot come to a contradiction off something nonexistent. This is as clear as I can formulate it.

N.B. In general it is not uncommon for < or > to change into $\leq$ or $\geq$, respectively when taking limits.

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Here's an example of a proof:

We have $f(x)\leq g(x)$ and $\lim\limits_{x\to a} f(x)=M$ and $\lim\limits_{x\to a} g(x)=N$. Now assume that $M > N$. The $\epsilon$-$\delta$ definition can be rephrased to say:

If $\lim\limits_{x\to a} f(x)=M$ then for every $\epsilon>0$ there is a $\delta >0$ such that $f(a-\delta;a+\delta)\subseteq (M-\epsilon; M+\epsilon)$.

And of course the same holds for $g(x)$ and $N$.

Now they say we may take any $\epsilon>0$, so we might as well take $\epsilon =\frac{M-N}{2}$. Now according to the definition there is a $\delta$ (convince yourself that one $\delta $ suffices) such that \begin{gather}f(a-\delta;a+\delta)\subseteq (M-\epsilon;M+\epsilon) \\\text{and}\\ g(a-\delta;a+\delta)\subseteq (N-\epsilon;N+\epsilon).\end{gather}

Now note that all numbers in $(M-\epsilon;M+\epsilon)$ are strictly greater than all numbers in $(N-\epsilon;N+\epsilon)$. This means that any $x\in(a-\delta;a+\delta)$ violates $f(x)\leq g(x)$, giving the sought for contradiction. Hence $M\leq N$.