If $\lim\limits_{x\to c} f(x)=L$, $\lim\limits_{x\to c} g(x)=M$, and $g(x)\geq f(x)$ for all $x\in\mathbb{R}$, then $M\geq L$.

Question

Prove that, if $\lim\limits_{x\to c} f(x)=L$, $\lim\limits_{x\to c} g(x)=M$, and $g(x)\geq f(x)$ for all $x\in\mathbb{R}$, then $M\geq L$.

So I have created a proof that seems to make sense for me but would like feedback letting me know if it's an acceptable proof. It goes as such:

Since $\lim\limits_{x\to c} f(x)=L$, then by the definition using epsilon-delta for all $\epsilon>0$ there exists $\delta_1>0$ s.t. $0<|x-c|<\delta_1$ implies $\big|f(x)-L\big|<\dfrac{\epsilon}{2}$. Since $\lim\limits_{x\to c} g(x)=M$, then again by definition for all $\epsilon>0$ there exists a $\delta_2>0$ s.t. $0<|x-c|<\delta_2$ implies $\big|g(x)-M\big|<\dfrac{\epsilon}{2}$.

Now, given $\epsilon>0$, choose $\delta=\min\left\{\delta_1,\delta_2\right\}$ so that we get the two results above $\big|f(x)-L\big|<\dfrac{\epsilon}{2}$ and $\big|g(x)-M\big|<\dfrac{\epsilon}{2}$. Definition of absolute value $\big|f(x)-L\big|<\dfrac{\epsilon}{2}$ implies $$-\dfrac{\epsilon}{2}+L< f(x) < \dfrac{\epsilon}{2}+L$$ and $\big|g(x)-M\big|<\dfrac{\epsilon}{2}$ similarly implies $$-\dfrac{\epsilon}{2}+M< g(x) < \dfrac{\epsilon}{2}+M\,.$$

Since $g(x)\geq f(x)$ for all $x$, then using the absolute value results above, we get $$-\dfrac{\epsilon}{2}+L< f(x) \leq g(x) < \dfrac{\epsilon}{2}+M\,,$$ whence $$-\dfrac{\epsilon}{2}+L < \dfrac{\epsilon}{2}+M\,,$$ which implies $M-L>-\epsilon$. Since $\epsilon>0$ that means $-\epsilon<0$ so $M-L\geq 0$ or $M\geq L$.

Answer 1

The last step is wrong.

$M-L>-\epsilon$ and $0>-\epsilon$ cannot give you the conclusion that $M-L\geq0$

It should be proved like this.

Suppose $M<L$ and define $\epsilon\equiv L-M$. Then you can get within that $\delta$ you defined, $$ g(x) < M + \epsilon/2 = L - \epsilon/2 < f(x) $$ which contradicts with $f(x)\leq g(x)$.

So the $M<L$ cannot happen, which means $L\geq M$.