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 $M\geq L$.

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    $\begingroup$ Your last conclusion is true because $\epsilon$ is arbitrary. $\endgroup$ Oct 15, 2018 at 21:10
  • $\begingroup$ Similar: math.stackexchange.com/q/2923693/403337 $\endgroup$
    – user403337
    Oct 15, 2018 at 21:22
  • $\begingroup$ Better replace your last step as $M-L>-\epsilon$ for every $\epsilon >0$ and hence $M-L\geq 0$. $\endgroup$
    – Paramanand Singh
    Oct 16, 2018 at 4:27

1 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 $L\geq M$.

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    $\begingroup$ Why can the conclusion not work? If at the beginning I let ε>0. So if M-L>-ε, we know ε>0 so -ε<0 meaning that for M-L to be greater than every single -ε which would get close to zero -1,-0.1,-0.001,-0.000001, etc. We would require M-L to be greater or equal to zero. $\endgroup$
    – user604647
    Oct 15, 2018 at 21:52
  • $\begingroup$ Your approach as well as the approach of @user604647 is correct. Think of it in this manner: "if a real number is greater than every negative real number, then it must be non-negative". $\endgroup$
    – Paramanand Singh
    Oct 16, 2018 at 4:24
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    $\begingroup$ @user604647 ok, now I understand your argument. It is correct. Just your writing makes your logic not very clear. You should end your proof like this. "In summary, $\forall \epsilon >0, M-L > -\epsilon. \Rightarrow M-L\geq 0$ $\endgroup$
    – MoonKnight
    Oct 16, 2018 at 6:36

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