Here's what I need to prove:
Suppose that $f(x)\le g(x)$ on some deleted neighborhood of the point $c$ and that $\lim_{x\to c} f(x)$ and $\lim_{x\to c} g(x)$ exist. Then $\lim_{x\to c} f(x) \le \lim_{x\to c} g(x)$.
The authors ask to prove it directly using the definition and not to use the fact that if $L > 0$, there is a $\delta >0$ such that $f(x)>0$ for $0<|x-c|<\delta$.
Here's my attempt:
If $f(x)\le g(x)$ on some deleted neighborhood of $c$ then there's a $\delta_{1} > 0$ such that $(c-\delta_{1} , c+ \delta_{1}) \setminus \{ c \}$ is contained in that neighborhood. Let $L_{1} := \lim_{x \to c } f(x)$ and Let $L_{2} := \lim_{x \to c } g(x)$. Assume that $L_{1} > L_{2}$. Now, there are $\delta_{2} > 0$ and $\delta_{3}>0$ such that $|f(x)-L_1| < L_1-L_2$ i.e. $L_2 < f(x)$ for $0<|x-c|< \delta_2$ and similarly $g(x) < L_1 $ for $0<|x-c|< \delta_{3}$. Now for $0<|x-c|< \min \{ \delta_1 , \delta_2 , \delta_3 \} $ we have $L_2 < f(x) \le g(x) < L_1 $ which contradicts our assumption.
Is my proof fine?
Let $L_1, L_2$ be the limits of $f,g$ respectively $\implies -\epsilon < f -L_1< \epsilon , -\epsilon <L_2-g < \epsilon\implies -2\epsilon < (f-g) + (L_2-L_1) < 2\epsilon\implies L_2-L_1> -2\epsilon +(g-f)\ge -2\epsilon$. Since this is true for any $\epsilon > 0$, we have: $L_2 - L_1 \ge 0\implies L_1 \le L_2$.