# 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 $M\geq L$.

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 $$M\geq L$$.

• Your last conclusion is true because $\epsilon$ is arbitrary. – hamam_Abdallah Oct 15 '18 at 21:10
• – Chris Custer Oct 15 '18 at 21:22
• Better replace your last step as $M-L>-\epsilon$ for every $\epsilon >0$ and hence $M-L\geq 0$. – Paramanand Singh Oct 16 '18 at 4:27

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 $$M and define $$\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 $$M cannot happen, which means $$L\geq M$$.

• 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. – user604647 Oct 15 '18 at 21:52
• 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". – Paramanand Singh Oct 16 '18 at 4:24
• @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, $\forall \epsilon >0, M-L > -\epsilon. \Rightarrow M-L\geq 0$ – MoonKnight Oct 16 '18 at 6:36