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.
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.
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?
Best Answer
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.