If $f(x) \le g(x)$ for all $x$ then $\lim_{x\to\infty} f(x) \le \lim_{x\to\infty} g(x)$.

Question

Given $f(x) \leq g(x)$ for all x and $\lim_{x \rightarrow \infty} f(x)=L_1$ , $\lim_{x \rightarrow \infty} g(x) = L_2 $ prove that $L_1\leq L_2$.

I tried to use a proof by contradiction but I got stuck as I'm unsure on how to use the definition of the limit here. Thanks for any help.

Answer 1

With limit definition, let $L_1>L_2$ so for $\varepsilon=\dfrac{L_1-L_2}{2}$ $$L_1-\dfrac{L_1-L_2}{2}<f(x)<L_1+\dfrac{L_1-L_2}{2}$$ where $x>M_1$ for a $M_1>0$ also $$L_2-\dfrac{L_1-L_2}{2}<g(x)<L_2+\dfrac{L_1-L_2}{2}$$ where $x>M_2$ for a $M_2>0$ , then if $x>\max\{M_1,M_2\}$ $$f(x)-g(x)>L_1-\dfrac{L_1-L_2}{2}-(L_2+\dfrac{L_1-L_2}{2})>0$$ which makes a contradiction with $f(x) \leq g(x)$ for all $x$.

Answer 2

Suppose $L_1>L_2$ and let $c=\frac{L_1+L_2}{2}$. For $x\gg0$, $f(x)>c>g(x)$.