Recall that the basic definition of the two-sided limit is as follows:

$\lim\limits_{x\to c} f(x)=L$ means that for every $\epsilon>0$, there exists a $\delta>0$, such that for every $x$, the expression $0< |x-c|<\delta$ implies $|f(x)-L|<\epsilon$. |

The proofs of the Limit Inequalities depend upon this definition of the limit.

- If $f(x)\le g(x)$ for all $x$ on the set $S=(a,c)\cup (c,b)$, $\lim\limits_{x\to c}f(x)=L$, and $\lim\limits_{x\to c}g(x)=M$, then $L\le M$.

We will assume that $L>M$, and show that this produces a contradiction. | Since the theorem provides two possible outcomes, and only excludes one outcome, we will show that the excluded outcome is not consistent with the given statements. So the excluded outcome is our beginning assumption. |

Choose $\epsilon=L-M$. | Since this is a proof by contradiction, we expect that our condition will no longer be true for every possible epsilon. Therefore, it will be sufficient to show that there exists one epsilon that produces the contradiction. This epsilon is positive, according to the assumption above. And we claim this is an epsilon that will result in the contradiction. |

Since $\lim\limits_{x\to c}[g(x)-f(x)]=M-L$ | By application of the Difference Law for limits. |

then there exists a $\delta>0$ such that for every $x$, the expression $0< |x-c| <\delta$ implies $|g(x)-f(x)-(M-L)|<\epsilon$. | By using the definition of the limit. |

Thus $-\epsilon < g(x)-f(x)-M+L < \epsilon$, $g(x)-f(x)-M+L < L-M$, |
We expanded the absolute value inequality, and then replaced $\epsilon$ by the value we chose for it above. |

$g(x)-f(x)<0$, and therefore $g(x) < f(x)$. |
Adding $M$ and subtracting $L$ from both sides removes them from the inequality. Then we added $f(x)$ to both sides. |

But this result contradicts the original hypothesis that $f(x)\le g(x)$, so the assumption must be false. Therefore, $L\le M$. | Having arrived at a contradiction, we find that the assumption was not consistent with the other statements. So we are forced to conclude that the assumption is never true under the given conditions, therefore the negation of the assumption must be true. And this concludes our proof. |

- If $f(x)\le g(x)\le h(x)$ for all $x$ on the set $S=(a,c)\cup (c,b)$, and $\lim\limits_{x\to c}f(x)=\lim\limits_{x\to c}h(x)=L$, then $\lim\limits_{x\to c}g(x)=L$.

Since $f(x)\le g(x)\le h(x)$ on set $S$, then $\lim\limits_{x\to c}f(x)\le\lim\limits_{x\to c}g(x)\le\lim\limits_{x\to c}h(x)$. | This is the result of the Limit Inequality Theorem. |

Thus $L\le\lim\limits_{x\to c}g(x)\le L$. | And the limits of the leftmost and rightmost expressions were part of the original hypothesis. |

Therefore $\lim\limits_{x\to c}g(x)=L$. | Since the limit was both less than or equal to $L$, and greater than or equal to $L$, the only conclusion is that the limit is equal to $L$. And thus our theorem has been proven. |

- If $f(x)=g(x)$ for all $x$ on the set $S=(a,c)\cup (c,b)$, then $\lim\limits_{x\to c}f(x)=\lim\limits_{x\to c}g(x)$.

Since $f(x)=g(x)$ on set $S$, then it is also true that $g(x)\le f(x)\le g(x)$. | Here, we are sandwiching $f(x)$ between two copies of $g(x)$. Since they are all equal, the sandwich is really quite thin, but still true. |

Therefore $\lim\limits_{x\to c}f(x)=\lim\limits_{x\to c}g(x)$. | Then we apply the Sandwich Theorem, and immediately obtain our conclusion. |

Strictly speaking, the All But One Point Theorem does not involve an inequality. However, we included it here because this proof does rely on the Sandwich Theorem.